首先之前的首先，我先宣布一下这个帖子作废了，但是还留着做纪念吧出新版了，大家可以看看哦~

首先，感谢@林海威大佬的“题解”让我有了思路
于是，我也决定来一波恐怖的题解！
时隔两年多，卷土重来！

啊这个帖子当我练练手吧……

😱😱😱

啊，对了，为了防止误导新人，先奉上一个非常正常的蒟蒻版代码：

#include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
} 

啊那个啥？

你不信都AC了？

自己测试去吧！

而且有些算法比直接A+B还要快几毫秒……

本“题解”参考了林海威大佬的“题解”，还有洛谷的部分“题解”
为什么加上删除线呢？因为这些都不是正经的题解

现在开始！

算法一、DFS一号

#include <bits/stdc++.h>
using namespace std;
int n = 2, a[5], s;
int dfs(int x, int sum) {
    if (x > n) return sum;
    int i = dfs(x + 1, sum);
    int j = dfs(x + 1, sum + a[x]);
    if (i == s) return i;
    if (j == s) return j;
    return -1;
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    cout << dfs(1, 0) << endl;
    return 0;
}

算法二、DFS二号

#include <bits/stdc++.h>
using namespace std;
int a, b;
int dfs(int x) {
    if (x <= 5) return x;
    return dfs(x / 2) + dfs(x - x / 2);
} 
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", dfs(a) + dfs(b));
    return 0;
}

算法三、BFS

#include <bits/stdc++.h>
using namespace std;
int n = 2, a[5], s;
queue<int> q;
void bfs() {
    q.push(0);
    int c = 0;
    while (q.size()) {
        c++;
        int f = q.front(); q.pop();
        if (f == s) {printf("%d\n", f); exit(0);}
        q.push(f + a[c]);
        q.push(f);
    }
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    bfs();
    return 0;
}

算法四、直接算咯

#include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
} 

算法五、二分

#include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    int l = 0, r = 200000000;
    while (l < r) {
        int mid = l + r >> 1;
        if (mid == a + b) {printf("%d\n", mid); return 0;}
        if (mid <  a + b) l = mid + 1;
        if (mid >  a + b) r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

算法六、稍微有点暴力的枚举

但是还是$1892ms$卡过了欸

#include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    for (int i = 0;i <= 200000000; i++) if (a + b == i) {printf("%d\n", i); break;}
    return 0;
} 

算法七、最短路之dijkstra

思路：定义节点1到节点2路径长度为a，节点2到节点3路径长度为b
则答案为节点1到节点3的最短路（也就是$a+b$）

#include <bits/stdc++.h>
using namespace std;
int w[5][5], d[5], v[5];
int n = 3;
void dijkstra() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            d[y] = min(d[y], d[x] + w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    dijkstra();
    printf("%d\n", d[3]);
    return 0;
}

算法八、最短路之SPFA

思路同上

#include <bits/stdc++.h>
using namespace std;
int a, b, n = 3;
int w[5][5], d[5], v[5];
queue<int> q;
void spfa() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0, v[1] = 1;
    q.push(1);
    while (q.size()) {
        int x = q.front(); q.pop();
        v[x] = 0;
        for (int i = 1;i <= n; i++) {
//          if (w[x][i] == 0x3f) continue;
            if (d[i] > d[x] + w[x][i]) {
                d[i] = d[x] + w[x][i];
                if (!v[i]) q.push(i), v[i] = 1;
            }
        }
    }
}
int main() {
    scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    spfa();
    printf("%d\n", d[3]);
    return 0;
}

算法九、最短路之Floyd

思路同上

#include <bits/stdc++.h>
using namespace std;
int d[5][5], n = 3;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(d, 0x3f, sizeof d);
    d[1][2] = a; d[2][3] = b;
    for (int k = 1;k <= n; k++)
        for (int i = 1;i <= n; i++)
            for (int j = 1;j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    printf("%d\n", d[1][3]);
    return 0;
}

算法十、高精

#include<bits/stdc++.h>
using namespace std;
string a0, b0;
int a[1005], b[1005];
int main(){
    cin >> a0 >> b0;
    int l1 = a0.size(), l2 = b0.size();
    for (int i = 0;i < l1; i++) a[l1 - i] = a0[i] - 48;
    for (int i = 0;i < l2; i++) b[l2 - i] = b0[i] - 48;
    l1 = max(l1, l2);
    for (int i = 1;i <= l1; i++) {
        a[i] += b[i];
        if (a[i] > 9) a[i + 1] += 1, a[i] %= 10;
    }
    if (a[max(l1, l2) + 1] > 0) l1++;
    for (int i = l1;i >= 1; i--) printf("%d", a[i]);
    return 0;
}

算法十一、最小生成树之kruskal

思路其实和最短路的一样，只是改成用最小生成树的方法求罢了

#include <bits/stdc++.h>
using namespace std;
struct rec {
    int x, y, z;
} edge[5];

int fa[5], m = 2, ans = 0;

int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
int cmp(rec a, rec b) { return a.z < b.z; }

int main() {
    int a, b; scanf("%d%d", &a, &b);
    edge[1] = (rec){1, 2, a};
    edge[2] = (rec){2, 3, b};
    for (int i = 1;i <= m + 1; i++) fa[i] = i;
    sort(edge + 1, edge + 1 + m, cmp);
    for (int i = 1;i <= m; i++) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if (x == y) continue;
        fa[x] = y;
        ans += edge[i].z;
    }
    printf("%d\n", ans);
    return 0;
}

算法十二、最小生成树之prim

思路同上

#include <bits/stdc++.h>
using namespace std;
int w[5][5], d[5], n = 3, ans, v[5];

void prim() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            if (!v[y]) d[y] = min(d[y], w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    prim();
    int ans = 0;
    for (int i = 2;i <= n; i++) ans += d[i];
    printf("%d\n", ans);
    return 0;
}

算法十三、前缀和

#include <bits/stdc++.h>
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 1;i <= 2; i++) scanf("%d", &a[i]), s[i] += a[i] + s[i - 1];
    printf("%d\n", s[2]);
    return 0;
}

算法十四、后缀和

#include <bits/stdc++.h>
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 2;i >= 1; i--) scanf("%d", &a[i]), s[i] += a[i] + s[i + 1];
    printf("%d\n", s[1]);
    return 0;
}

算法十五、位运算

#include <bits/stdc++.h>
using namespace std;
int add(int a, int b) {
    if (b == 0) return a;
    return add(a ^ b, (a & b) << 1);
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d\n", add(a, b));
    return 0;
}

算法十六、树的直径——BFS

emmm……思路继续和最短路的一样。。。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;

int head[maxn * 2],edge[maxn * 2],Next[maxn * 2],ver[maxn * 2];
int vis[maxn], dist[maxn];
int n = 3, p, q, d;
int tot = 0;
int maxd = 0;

void add(int u,int v,int w) {
    ver[ ++ tot] = v,edge[tot] = w;
    Next[tot] = head[u],head[u] = tot;
}

int BFS(int u) {
    queue<int>Q;
    while(!Q.empty()) Q.pop();
    memset(vis, 0, sizeof vis);
    memset(dist, 0, sizeof dist);  
    Q.push(u);
    int x, max_num = 0;
    while(!Q.empty()) {
        x = Q.front();
        Q.pop();
        vis[x] = 1;
        for(int i = head[x]; i ; i = Next[i]) {
            int y = ver[i];
            if(vis[y]) continue;
            vis[y] = 1;
            dist[y] = dist[x] + edge[i];
            if(dist[y] > maxd) {
                maxd = dist[y];
                max_num = y;
            }
            Q.push(y);
        }
    }
    return max_num;
}
int main(void) {
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    BFS(BFS(1));
    int ans = 0;
    for (int i = 1;i <= n; i++) ans = max(ans, dist[i]);
    printf("%d\n", ans);
    return 0;
}

算法十七、树的直径——DFS

思路同上

#include<bits/stdc++.h>
#define maxn 100000
using namespace std;
struct node {
    int u, v, w, nex;
} edge[2 * maxn + 10];
int n = 3, d[maxn + 10], head[maxn + 10], f_num, cnt = 0, ans;
inline void add(int x,int y,int z) {
    edge[++cnt].u = x;
    edge[cnt].v = y;
    edge[cnt].w = z;
    edge[cnt].nex = head[x];
    head[x] = cnt;
}
inline void dfs(int x, int fa) {
    if(ans < d[x]) {
        ans = d[x];
        f_num = x;
    }
    for (int i = head[x]; i != -1; i = edge[i].nex) {
        int j = edge[i].v;
        if (j == fa)continue;
        d[j] = d[x] + edge[i].w;    
        dfs(j, x);
    }
}
int main() {
    memset(head, -1, sizeof(head));
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dfs(1, 0);
    ans = 0;
    d[f_num] = 0;
    dfs(f_num, 0);
    for (int i = 1;i <= n; i++) ans = max(ans, d[i]);
    printf("%d", ans);
    return 0;
}

算法十八、树的直径——树形DP

还是一样咯

#include <bits/stdc++.h>
using namespace std;
int f[5], n = 3, cnt, h[5], ans, dis[5];
struct edge {
    int to, next, vi;
} e[5];
void add(int u, int v, int w) {
    e[cnt].to= v;
    e[cnt].vi = w;
    e[cnt].next = h[u];
    h[u] = cnt++;
}
void dp(int u, int fa) {
    for (int i = h[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if (v == fa) continue;
        dp(v, u);
        ans = max(ans, dis[v] + dis[u] + e[i].vi);
        dis[u] = max(dis[u], dis[v] + e[i].vi);
    }
}
int main() {
    memset(h, -1, sizeof h);
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dp(1, 0);
    printf("%d\n", ans);
    return 0;
}

算法十九、网络流

从别的大佬那边搞过来的，但是一点都看不懂┭┮﹏┭┮

#include<bits/stdc++.h>
using namespace std;
#define set(x) Set(x)
#define REP(i,j,k) for (int i=(j),_end_=(k);i<=_end_;++i)
#define DREP(i,j,k) for (int i=(j),_start_=(k);i>=_start_;--i)
#define mp make_pair
#define x first
#define y second
#define pb push_back
template<typename T> inline bool chkmin(T &a,const T &b){ return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a,const T &b){ return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef pair<int,int> node;
const int dmax = 1010, oo = 0x3f3f3f3f;
int n, m;
int a[dmax][dmax] , ans;
int d[dmax], e[dmax];
priority_queue <node> q;
inline bool operator >(node a,node b){ return a.y>b.y; }
bool p[dmax];
void Set(int x){ p[x] = 1; }
void unset(int x){ p[x] = 0; }
bool check(int x){ return x != 1 && x != n && !p[x] && e[x] > 0; }
void preflow(){
    e[1] = oo;
    d[1] = n - 1;
    q.push(mp(1, n - 1));
    set(1);
    while (!q.empty()) {
        bool flag = 1;
        int k = q.top().x;
        q.pop(), unset(k);
        DREP(i, n, 1)
        if ((d[k] == d[i] + 1 || k == 1) && a[k][i] > 0){
            flag = 0;
            int t = min(a[k][i], e[k]);
            e[k] -= t;
            a[k][i] -= t;
            e[i] += t;
            a[i][k] += t;
            if (check(i)) {
                q.push(mp(i, d[i]));
                set(i);
            }
            if (e[k] == 0) break;
        }
        if (flag) {
            d[k] = oo;
            REP(i, 1, n)
            if (a[k][i] > 0) chkmin(d[k], d[i] + 1);
        }
        if (check(k)) {
            q.push(mp(k, d[k]));
            set(k);
        }
    }
    ans = e[n];
}
int main() {
    n = 2, m = 2;
    int x, y;
    scanf("%d%d", &x, &y);
    a[1][2] += x + y;
    preflow();
    printf("%d\n", ans);
    return 0;
}

算法二十、线段树

转化为区间求和问题

#include <bits/stdc++.h>
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
using namespace std;
struct SegmentTree {
    int l, r; //区间左右端点 
    long long sum, add; //sum 区间和  add 延迟标记 
} tree[400010];
int a[100010], n = 1, m = 2;
void build (int p, int l, int r) {
    l(p) = l, r(p) = r;
    if(l == r) {sum(p) = a[l]; return;}
    int mid = l + r >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
void spread(int p) {
    if(add(p)) { //节点p有标记 
        sum(p * 2) += add(p) * (r(p * 2) - l(p * 2) + 1); //更新左子节点信息 
        sum(p * 2 + 1) += add(p) * (r(p * 2 + 1) - l(p * 2 + 1) + 1); //更新右子节点
        add(p * 2) += add(p); //给左子节点打延迟标记 
        add(p * 2 + 1) += add(p); //给右子节点打延迟标记 
        add(p) = 0; //清除p的标记 
    }
}
void change(int p, int l, int r, int d) {
    if(l <= l(p) && r >= r(p)) { //完全覆盖 
        sum(p) += (long long) d * (r(p) - l(p) + 1); //更新节点信息 
        add(p) += d; //给节点打延迟标记 
        return;
    }
    spread(p); //下传延迟标记 
    int mid = l(p) + r(p) >> 1;
    if(l <= mid) change(p * 2, l, r, d);
    if(r > mid) change(p * 2 + 1, l, r, d);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
long long ask(int p, int l, int r) {
    if(l <= l(p) && r >= r(p)) return sum(p);
    spread(p);
    int mid = l(p) + r(p) >> 1;
    long long val = 0;
    if(l <= mid) val += ask(p * 2, l, r);
    if(r > mid) val += ask(p * 2 + 1, l, r);
    return val;
}
int main() {
    a[1] = 0;
    build(1, 1, n);
    while(m--) { 
        int d = 0;
        scanf("%d", &d);
        change(1, 1, 1, d);
    }
    printf("%lld\n", ask(1, 1, 1));
    return 0;
}

算法二十一、树状数组

思路一样，区间求和

#include<bits/stdc++.h>
using namespace std;
const int SIZE = 100010;
int a[SIZE], n = 1, m = 2;
long long c[2][SIZE], sum[SIZE];

long long ask(int k, int x) {
    long long ans = 0;
    for(; x ; x -= x & -x) ans += c[k][x];
    return ans;
}

void add(int k,int x,int y) {
    for(; x <= n; x += x & -x) c[k][x] += y;
}

int main() {
    a[1] = 0;
    while(m--) {
        int d = 0;
        scanf("%d", &d);
        add(0, 1, d);
        add(0, 2, -d);
        add(1, 1, d);
        add(1, 2, -2 * d);
    }
    long long ans = sum[1] + 2 * ask(0, 1) - ask(1, 1);
    ans -= sum[0] + 1 * ask(0, 0) - ask(1, 0);
    printf("%lld\n", ans);
    return 0;
}

算法二十二、分块

思路一样，区间求和

#include<bits/stdc++.h>
using namespace std;
long long a[50000010], sum[50000010], add[50000010];
int L[50000010], R[50000010];
int pos[50000010];
int n = 1, m = 2, t;

void change(int l, int r, long long d) {
    int p = pos[l], q = pos[r];
    if (p == q) {
        for (int i = l; i <= r; i++) a[i] += d;
        sum[p] += d*(r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++) add[i] += d;
        for (int i = l; i <= R[p]; i++) a[i] += d;
        sum[p] += d*(R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) a[i] += d;
        sum[q] += d*(r - L[q] + 1);
    }
}

long long ask(int l, int r) {
    int p = pos[l], q = pos[r];
    long long ans = 0;
    if (p == q) {
        for (int i = l; i <= r; i++) ans += a[i];
        ans += add[p] * (r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++)
            ans += sum[i] + add[i] * (R[i] - L[i] + 1);
        for (int i = l; i <= R[p]; i++) ans += a[i];
        ans += add[p] * (R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) ans += a[i];
        ans += add[q] * (r - L[q] + 1);
    }
    return ans;
}

int main() {
    a[1] = 0;
    t = sqrt(n*1.0);
    for (int i = 1; i <= t; i++) {
        L[i] = (i - 1)*sqrt(n*1.0) + 1;
        R[i] = i*sqrt(n*1.0);
    }
    if (R[t] < n) t++, L[t] = R[t - 1] + 1, R[t] = n;
    for (int i = 1; i <= t; i++)
        for (int j = L[i]; j <= R[i]; j++) {
            pos[j] = i;
            sum[i] += a[j];
        }
    while (m--) {
        int d;
        scanf("%d", &d);
        change(1, 1, d);
    }
    printf("%lld\n", ask(1, 1));
}

算法二十三、LCT

嘿嘿(●ˇ∀ˇ●)
洛谷加油！洛谷加油！
啊又是洛谷来的快递。。。我一点都看不懂┭┮﹏┭┮

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int data,rev,sum;
    node *son[2],*pre;
    bool judge();
    bool isroot();
    void pushdown();
    void update();
    void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x)
{
    node *now=lct+ ++top;
    now->data=x;
    now->pre=now->son[1]=now->son[0]=lct;
    now->sum=0;
    now->rev=0;
    return now;
}
bool node::judge()
{
    return pre->son[1]==this;
}
bool node::isroot()
{
    if(pre==lct)return true;
    return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown()
{
    if(this==lct||!rev)return;
    swap(son[0],son[1]);
    son[0]->rev^=1;
    son[1]->rev^=1;
    rev=0;
}
void node::update()
{
    sum=son[1]->sum+son[0]->sum+data;
}
void node::setson(node *child,int lr)
{
    this->pushdown();
    child->pre=this;
    son[lr]=child;
    this->update();
}
void rotate(node *now)
{
    node *father=now->pre,*grandfa=father->pre;
    if(!father->isroot()) grandfa->pushdown();
    father->pushdown();
    now->pushdown();
    int lr=now->judge();
    father->setson(now->son[lr^1],lr);
    if(father->isroot()) now->pre=grandfa;
    else grandfa->setson(now,father->judge());
    now->setson(father,lr^1);
    father->update();
    now->update();
    if(grandfa!=lct) grandfa->update();
}
void splay(node *now)
{
    if(now->isroot())return;
    for(; !now->isroot(); rotate(now))
    if(!now->pre->isroot())
    now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now)
{
    node *last=lct;
    for(; now!=lct; last=now,now=now->pre) {
        splay(now);
        now->setson(last,1);
    }
    return last;
}
void changeroot(node *now)
{
    access(now)->rev^=1;
    splay(now);
}
void connect(node *x,node *y)
{
    changeroot(x);
    x->pre=y;
    access(x);
}
void cut(node *x,node *y)
{
    changeroot(x);
    access(y);
    splay(x);
    x->pushdown();
    x->son[1]=y->pre=lct;
    x->update();
}
int query(node *x,node *y)
{
    changeroot(x);
    node *now=access(y);
    return now->sum;
}
int main()
{
    scanf("%d%d",&a,&b);
    node *A=getnew(a);
    node *B=getnew(b);
    connect(A,B);
    cut(A,B);
    connect(A,B);
    printf("%d",query(A,B));
    return 0;
}

算法二十四、Splay

洛谷加油！

#include <bits/stdc++.h>
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
    sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
    if(rev[x]){
        swap(ch[x][0], ch[x][1]);
        if(ch[x][1]) rev[ch[x][1]] ^= 1;
        if(ch[x][0]) rev[ch[x][0]] ^= 1;
        rev[x] = 0;
    }
    if(tag[x]){
        if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
        if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
        tag[x] = 0;
    }
}
void rotate(int x, int &k){
    int y = fa[x], z = fa[fa[x]];
    int kind = ch[y][1] == x;
    if(y == k) k = x;
    else ch[z][ch[z][1]==y] = x;
    fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
    ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
    push_up(y); push_up(x);
}
void splay(int x, int &k){
    while(x != k){
        int y = fa[x], z = fa[fa[x]];
        if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
        else rotate(y, k);
        rotate(x, k);
    }
}
int kth(int x, int k){
    push_down(x);
    int r = sz[ch[x][0]]+1;
    if(k == r) return x;
    if(k < r) return kth(ch[x][0], k);
    else return kth(ch[x][1], k-r);
}
void split(int l, int r){
    int x = kth(rt, l), y = kth(rt, r+2);
    splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
    split(l, r);
    rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
    split(l, r);
    tag[ch[ch[rt][1]][0]] += v;
    val[ch[ch[rt][1]][0]] += v;
    push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
    if(l > r) return 0;
    if(l == r){
        fa[l] = f;
        sz[l] = 1;
        return l;
    }
    int mid = l + r >> 1;
    ch[mid][0] = build(l, mid-1, mid);
    ch[mid][1] = build(mid+1, r, mid);
    fa[mid] = f;
    push_up(mid);
    return mid;
}
int asksum(int l, int r){
    split(l, r);
    return sum[ch[ch[rt][1]][0]];
}
int main(){
    //总共两个数
    n = 2;
    rt = build(1, n+2, 0);//建树
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        add(i, i, x);//区间加
    }
    rever(1, n);//区间翻转
    printf("%d\n", asksum(1, n));//区间求和
    return 0;
}

算法二十五、LCA

洛谷加油！

#include<cstdio>                                                  //头文件
#define NI 2                                                          
//从来不喜欢算log所以一般用常数 不知道算不算坏习惯 因为3个节点 所以log3(当然以2为底)上取整得2
struct edge
{
    int to,next,data;                                              //分别表示边的终点，下一条边的编号和边的权值
}e[30];                                                                     //邻接表，点少边少开30是为了浪啊
int v[10],d[10],lca[10][NI+1],f[10][NI+1],tot=0;      //数组开到10依然为了浪
//数组还解释嘛，v表示第一条边在邻接表中的编号，d是深度，lca[x][i]表示x向上跳2^i的节点，f[x][i]表示x向上跳2^i的距离和
void build(int x,int y,int z)                                      //建边
{
    e[++tot].to=y; e[tot].data=z; e[tot].next=v[x]; v[x]=tot;
    e[++tot].to=x; e[tot].data=z; e[tot].next=v[y]; v[y]=tot;
}
void dfs(int x)                                                        //递归建树
{
    for(int i=1;i<=NI;i++)                                   //懒，所以常数懒得优化
        f[x][i]=f[x][i-1]+f[lca[x][i-1]][i-1],
        lca[x][i]=lca[lca[x][i-1]][i-1];                   //建树的同时进行预处理
    for(int i=v[x];i;i=e[i].next)                              //遍历每个连接的点
    {
        int y=e[i].to;
        if(lca[x][0]==y) continue;
        lca[y][0]=x;                                       //小技巧：lca[x][0]即为x的父亲~~（向上跳2^0=1不就是父节点嘛）
        f[y][0]=e[i].data;
        d[y]=d[x]+1;
        dfs(y);                                            //再以这个节点为根建子树【这里真的用得到嘛？？】
    }
}
int ask(int x,int y)                                             //询问，也是关键
{                                                                        
    if(d[x]<d[y]) {int t=x;x=y;y=t;}                  //把x搞成深的点
    int k=d[x]-d[y],ans=0;
    for(int i=0;i<=NI;i++)
        if(k&(1<<i))                                      //若能跳就把x跳一跳
            ans+=f[x][i],                              //更新信息
            x=lca[x][i];
    for(int i=NI;i>=0;i--)                                  //不知道能不能正着循环，好像倒着优，反正记得倒着就好了
        if(lca[x][i]!=lca[y][i])                            //如果x跳2^i和y跳2^j没跳到一起就让他们跳
            ans+=f[x][i]+f[y][i],
            x=lca[x][i],y=lca[y][i];
    return ans+f[x][0]+f[y][0];                           //跳到LCA上去（每步跳的时候都要更新信息，而且要在跳之前更新信息哦~）
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    build(1,2,a);
    build(1,3,b);                                                       //分别建1 2、1 3之间的边
    dfs(1);                                                                //以1为根建树
    printf("%d",ask(2,3));                                         //求解2 3到它们的LCA的距离和并输出
}

算法二十六、字典树

洛谷继续加油！

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i<strlen(str1);i++)
    {
         if(str1[i]=='-'){
             f1=true;continue;
         }
         if(!s[t].str[str1[i]-'0'])
         s[t].str[str1[i]-'0']=++tot;
         t=s[t].str[str1[i]-'0'];
         s[t].sum=str1[i]-'0';
    }
}
int query()
{
   int t=0;int s1=0;
   for(int i=0;i<strlen(str1);i++)
   {
           if(str1[i]=='-') continue;
           if(!s[t].str[str1[i]-'0']) return s1;
           t=s[t].str[str1[i]-'0'];
           s1=s1*10+s[t].sum;
   }
   return s1;
}
int main()
{    
  for(int i=1;i<=2;i++)
  {
      f1=false;
      scanf("%s",str1);
    built();
    if(f1)
      ss-=query();
      else ss+=query();
  }
  printf("%d",ss);
  return 0;    
}

上面的26个解法都AC了（我测试过~你们不信可以自己试试去！）

接下来的解法不太正经。。。有些解法和那个特别正经的蒟蒻版直接算A+B没啥区别（根本没有区别）就是改了点格式。。。

小小声：都是从别人那里搞过来的奇葩代码。。。

还会陆续补充~

一、非常标准的绕圈代码

#include <bits/stdc++.h>
int __________=45,___________=5;
template<typename _>bool ________(_ __) {return __>=(__________+3)&&__<=(__________+12);}
template<typename _>void ___(_ __){if(__<(___________-5))__=abs(__),putchar('-');
if (__>(___________+4))___(__/(___________+5));putchar (__%(___________+5)+(__________+3));}
template<typename _>void ____(_ &__){__=(___________-5);_ _____=(__________-___________*9),______=getchar();
while(!________(______))_____^=!(______^__________),______=getchar();
while(________(______))__=(__<<1)+(__<<3)+(______&(__________-30)),______=getchar(); __=_____?-__:__;}
int main() {int _, __; ____(_); ____(__); ___(_+__);}

二、函数解法

#include <bits/stdc++.h>
using namespace std;
int add(int a, int b) { return a + b; }
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d\n", add(a, b));
    return 0;
}

三、结构体解法

#include <bits/stdc++.h>
using namespace std;
struct s{
    int a, b;
}x;
int main() {
    scanf("%d%d", &x.a, &x.b);
    printf("%d\n", x.a + x.b);
    return 0;
}

四、打表解法

这个测试数据只有@yxc大佬知道。。。
所以只有yxc大佬能用这个解法。。。
不过像y总这样的人生赢家肯定不会用这种可耻的打表的

但是

你们知道有些人打表写了3047行么？？？

就在$2021.10.6$的晚上$18:42$分
有个可耻的人私聊我
说他找到了A+B的打表解法
啊给大家分享一下这个$len$的可耻！

大家说这个人是不是闲的没事干！？！？

#include <bits/stdc++.h>
using namespace std;
int a, b; int main() { 
    scanf("%d%d", &a, &b);
    if (a == 3 && b == 4) printf("7");
    if (a == 45 && b == 55) printf("100");
    if (a == 123 && b == 321) printf("444");
    if (a == 91086199 && b == 18700332) printf("109786531");
    if (a == 42267194 && b == 60645282) printf("102912476");
    if (a == 69274392 && b == 10635835) printf("79910227");
    if (a == 5710219 && b == 85140568) printf("90850787");
    if (a == 75601477 && b == 24005804) printf("99607281");
    if (a == 70597795 && b == 90383234) printf("160981029");
    if (a == 82574652 && b == 22252146) printf("104826798");
    return 0;           //hh，这个len没加上return 0，还是我加的……
}

五、恐怖解法（AC一个点的成功率都≈0）

#include <bits/stdc++.h>
using namespace std;
int main() {
    srand((unsigned)time(NULL));
    int a = rand() % 100000000;
    printf("%d\n", a);
    return 0;
}

六、暴力$^2$解法

#include <bits/stdc++.h>
using namespace std;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    for (int i = 1;i <= 100000000; i++)
        for (int j = 1;j <= 100000000; j++)
            if (a == i && b == j) printf("%d\n", i + j);
    return 0;
}