A

#include <iostream>

using namespace std;

int cal(int x){
    int res = 0;
    while(x) res += (x % 10 == 2), x /= 10;
    return res;
}
int main(){
    int res = 0;
    for(int i = 1;i <= 2020;i ++ )
        res += cal(i);

    printf("%d\n", res);

    return 0;
}

B

#include <iostream>

using namespace std;

const int N = 2021;

int gcd(int a, int b){
    return b == 0 ? a : gcd(b, a % b);
}

int main(){
    int res = 0;
    for(int i = 1;i < N;i ++ )
        for(int j = 1;j < N;j ++ )
            if(gcd(i, j) == 1) res ++ ;

    printf("%d\n", res);

    return 0;
}

C

找规律

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
    int delta = 4;
    int start = 1;
    for(int i = 1;i < 20;i ++ ){
        start += delta;
        delta += 4;
    }
    cout << start << endl;

    return 0;
}

暴力模拟

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 1000;

int g[N][N];

bool check(int x){
    if(x <= 0) return false;
    return true;
}
int main(){

    int x = 1, y = 1;
    int dir = -1, now = 1;

    for(int i = 1;i <= 10000;){
        g[x][y] = i ++ ;
        // cout << g[x][y] << endl;
        int nx = x + dir, ny = y - dir;
        if(!check(nx) || !check(ny)){
            dir *= -1;
            if(!check(nx)) nx = 1, ny = ++ now;
            else ny = 1, nx = ++ now;
        }
        x = nx, y = ny;
    }

    for(int i = 1;i <= 20;i ++ ){
        for(int j = 1;j <= 20;j ++ )
            cout << g[i][j] << ' ';
        cout << endl;
    }
    return 0;
}

D

暴力模拟就行

E

繁琐建图，然后二进制枚举，最后 dfs 或者 bfs check 是否合法就行

F

#include <iostream>
#include <cmath>

using namespace std;

int good, pass;

int main(){

    int n; scanf("%d", &n);
    for(int i = 1;i <= n;i ++ ){
        int x; scanf("%d", &x);
        if(x >= 60) good ++ ;
        if(x >= 85) pass ++ ;
    }
    printf("%.0lf\%\n", round(good * 1.0 / n * 100));
    printf("%.0lf\%\n", round(pass * 1.0 / n * 100));

    return 0;
}

G

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 10010, base = 10000;

int n;
string ans, res;

int dir_f[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int dir_s[23] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool check(int year){
    if(year % 400 == 0 || year % 4 == 0 && year % 100 != 0) return true;
    return false;
}
bool check(string now, string date){
    int year = stoi(now);
    int month = stoi(date.substr(0, 2));
    int day = stoi(date.substr(2));
    if(month > 12 || month <= 0) return false;
    if(day <= 0 || day > 31) return false;
    if(check(year)){
        if(dir_s[month - 1] >= day) return true;
        else return false;
    }else{
        if(dir_f[month - 1] >= day) return true;
        else return false;
    }
}

bool check(string now){
    string year = now.substr(0, 4);
    string date = now.substr(4);
    if(!check(year, date)) return false;
    char a = now[0], b = now[1];
    if(now[2] != a || now[5] != a || now[7] != a) return false;
    if(now[3] != b || now[4] != b || now[6] != b) return false;
    return true;
}

int main(){

    scanf("%d", &n);
    for(int i = n / base + 1;;i ++ ){
        string now = to_string(i);
        string another = now; reverse(another.begin(), another.end());
        // cout << now << endl;
        if(check(now, another)){
            ans = now + another;
            break;
        }
    }

    for(int i = n / base + 1;;i ++ ){
        string now = to_string(i);
        string another = now; reverse(another.begin(), another.end());
        now += another;
        // cout << now << endl;
        if(check(now)){
            res = now;
            break;
        }
    }

    cout << ans << endl;
    cout << res << endl;

    return 0;
}

H

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 100010;

// last 记录每个字母最后一个位置出现是在哪里
char s[N];
int last[26];
int f[N];

int main(){
    scanf("%s", s + 1); int n = strlen(s + 1);
    f[0] = 0;
    for(int i = 1;i <= n;i ++ ){
        f[i] += f[i - 1];
        f[i] += i - last[s[i] - 'a'];
        last[s[i] - 'a'] = i;
    }
    for(int i = 1;i <= n;i ++ ) f[i] += f[i - 1];
    cout << f[n] << endl;

    return 0;
}

I

每次增加交点加一个平面