快读,快写 
inline int read(){
    int x=0;bool f=0;char c=getchar();
    while (c<'0'||c>'9'){if (c=='-')f=1;c=getchar();}
    while (c>='0'&&c<='9'){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
inline void print(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) print(x/10);
    putchar(x%10+'0');   
} 

超级快读测试代码(数据量极大,比关闭同步流快)
#include<bits/stdc++.h>
#define int long long 
#define endl '\n'
using namespace std;
struct fio
{
    inline char gc()
    {
        static const int IN_LEN = 1 << 18 | 1;
        static char buf[IN_LEN], *s, *t;
        static streambuf *sb = cin.rdbuf();
        return (s == t) && (t = (s = buf) + sb->sgetn(buf, IN_LEN)), s == t ? -1 : *s++;
    }
    inline fio &operator>>(string &s)
    {
        static char ch;
        ch = gc();
        s.clear();
        for (; isspace(ch); ch = gc())
            if (ch == -1)
                return *this;
        for (; !isspace(ch) && ch != -1; ch = gc())
            s += ch;
        return *this;
    }
    inline fio &operator>>(char *s)
    {
        static char ch;
        ch = gc();
        for (; isspace(ch); ch = gc())
            if (ch == -1)
                return *this;
        for (; !isspace(ch) && ch != -1; ch = gc())
            *(s++) = ch;
        return *this;
    }
    inline void gtline(string &s)
    {
        static char ch;
        ch = gc();
        s.clear();
        for (; ch != '\n' && ch != -1; ch = gc())
            s += ch;
    }
    template <typename T>
    inline fio &operator>>(vector<T> &n)
    {
        for (auto &it : n)
        {
            *this >> it;
        }
        return *this;
    }
    template <typename T>
    inline fio &operator>>(T &n)
    {
        static char ch, sgn;
        ch = gc(), sgn = 0;
        for (; !isdigit(ch); ch = gc())
        {
            if (ch == -1)
                return *this;
            sgn |= ch == '-';
        }
        for (n = 0; isdigit(ch); ch = gc())
            n = n * 10 + (ch ^ '0');
        if (ch == '.')
        {
            T tmp = 0;
            ch = gc();
            for (; isdigit(ch); ch = gc())
                tmp = tmp * 10 + (ch ^ '0');
            while (tmp > 1)
                tmp /= 10;
            n += tmp;
        }
        sgn && (n = -n);
        return *this;
    }
} io;
//读入用io>>n;
//输出还是用cout<<n 
signed main()
{
    int T;
    io>>T;
    while(T--)
    {
        int n;
        io>>n;
        cout<<n<<endl;
    }
}

版本2
struct I {
    char fin[_], *p1 = fin, *p2 = fin;
    inline char gc() {
        return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
    }
    inline I& operator>>(int& x) {
        bool sign = 1;
        char c = 0;
        while (c < 48) ((c = gc()) == 45) && (sign = 0);
        x = (c & 15);
        while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
        x = sign ? x : -x;
        return *this;
    }
    inline I& operator>>(double& x) {
        bool sign = 1;
        char c = 0;
        while (c < 48) ((c = gc()) == 45) && (sign = 0);
        x = (c - 48);
        while ((c = gc()) > 47) x = x * 10 + (c - 48);
        if (c == '.') {
            double d = 1.0;
            while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
        }
        x = sign ? x : -x;
        return *this;
    }
    inline I& operator>>(char& x) {
        do
            x = gc();
        while (isspace(x));
        return *this;
    }
    inline I& operator>>(string& s) {
        s = "";
        char c = gc();
        while (isspace(c)) c = gc();
        while (!isspace(c) && c != EOF) s += c, c = gc();
        return *this;
    }
} in;
struct O {
    char st[100], fout[_];
    signed stk = 0, top = 0;
    inline void flush() {
        fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
    }
    inline O& operator<<(int x) {
        if (top > (1 << 20)) flush();
        if (x < 0) fout[top++] = 45, x = -x;
        do
            st[++stk] = x % 10 ^ 48, x /= 10;
        while (x);
        while (stk) fout[top++] = st[stk--];
        return *this;
    }
    inline O& operator<<(char x) {
        fout[top++] = x;
        return *this;
    }
    inline O& operator<<(string s) {
        if (top > (1 << 20)) flush();
        for (char x : s) fout[top++] = x;
        return *this;
    }
} out;

1
[3, 1, 4, 1, 5, 9, 2, 6]
2 4
int t; cin>>t; getchar();
while(t--){
    string s;
    getline(cin,s);
    int x,y; cin>>x>>y;
    getchar();
}

二分:最小值最大,最大值最小,单调

求区间合并后的长度(没验证)
n是区间个数，m是所有区间总的范围
int CountSize(){
    int flg[...], cnt = 0, ret = 0;
    for (int i = 0; i < n; i++){
        flg[ left[i] ]++;
        flg[ right[i]+1 ]--;
    }
    for (int i = 0; i < m; i++){
        cnt += flg[i];
        if (cnt > 0){
            ret++;
        }
    }
    return ret;
}

归并排序求正序数
int n; int ans; int a[N]; int tmp[N];
void merge(int q[], int l, int r){
    if (l >= r) return;
    int mid = l + r >> 1;
    merge(q, l, mid);
    merge(q, mid + 1, r);
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r){
        if (q[i] <= q[j]){
            tmp[k ++ ] = q[i ++ ];
            ans=(ans+r-j+1)%mod;
        } 
        else tmp[k ++ ] = q[j ++ ];
    }
    while (i <= mid) tmp[k ++ ] = q[i ++ ];
    while (j <= r) tmp[k ++ ] = q[j ++ ];

    for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}
主函数: 
merge(a,1,n);
归并排序求逆序对数 
void merge(int q[], int l, int r){
    if (l >= r) return;
    int mid = l + r >> 1;
    merge(q, l, mid);
    merge(q, mid + 1, r);
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r){
        if (q[i] <= q[j]){
            tmp[k ++ ] = q[i ++ ];
        } 
        else{
            tmp[k ++ ] = q[j ++ ];
            ans=ans+mid-i+1;
        }
    }
    while (i <= mid) tmp[k ++ ] = q[i ++ ];
    while (j <= r) tmp[k ++ ] = q[j ++ ];

    for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}

进制转换
输入格式:共三行,第一行是一个正整数,表示需要转换的数的进制n(2≤n≤16),第二行是一个n进制数,若n>10则用大写字母A−F表示数码10−15,并且该n进制数对应的十进制的值不超过1000000000,第三行也是一个正整数,表示转换之后的数的进制m(2≤m≤16)
输出格式:一个正整数，表示转换之后的m进制数
输入:16 FF 2 输出:11111111
int b,m; char s[40],t[40];
void upper(char c[]){ for(int i=0;i<40;i++) if(c[i]>='a'&&c[i]<='z') c[i]+='A'-'a';}
signed main(){ cin>>b>>s>>m; itoa(strtol(s,NULL,b),t,m); upper(t); cout<<t; }