A - Find Multiple

#include<bits/stdc++.h>
using namespace std;

int main(){
   
    ios::sync_with_stdio(0);
    int a, b, c;
    cin >> a >> b >> c;
    for (int i = a; i <= b;i++){
   
        if(i%c==0){
   
            cout << i;
            return 0;
        }
    }
    cout << -1;
    return 0;
}

B - Base K


一开始没开long long WA了一发，我***


#include<bits/stdc++.h>
using namespace std;
int k;
long long f(string s){
   
    long long num = 0, m = 1;
    for (int i = s.size() - 1; i >= 0;i--){
   
        num += (s[i] - '0') * m;
        m *= k;
    }
    return num;
}
int main(){
   
    ios::sync_with_stdio(0);
    cin >> k;
    string a, b;
    cin >> a >> b;
    cout << f(a) * f(b);
    return 0;
}

C - Long Sequence


#include<bits/stdc++.h>
using namespace std;

int main(){
   
    ios::sync_with_stdio(0);
    long long n, x, sum=0;
    long long a[100100];
    cin >> n;
    for (int i = 0; i < n; i++){
   
        cin >> a[i];
        sum += a[i];
    }
    //cout << sum;//10
    cin >> x;
    long long m;
    m = x / sum;
    x = x - sum * m;
    long long num=0,cnt=0;
    for (int i = 0; i < n;i++){
   
        if(num>x){
   
            break;
        }
        num += a[i];
        cnt++;
    }
    cout << n * m + cnt;
    return 0;
}

D - FG operation


dp数组开大必TLE


#include<bits/stdc++.h>
using namespace std;
#define mod 998244353
int main(){
   
    ios::sync_with_stdio(0);
    int n;
    int a[100100],dp[100010][12];
    for (int i = 1; i <= n;i++){
   
        cin >> a[i];
    }
    dp[1][a[1]] = 1;
    for (int i = 1; i <= n;i++){
   
        for (int j = 0; j < 10;j++){
   
            if(dp[i][j]){
   
                dp[i + 1][(j + a[i + 1]) % 10] += dp[i][j];
                dp[i + 1][(j + a[i + 1]) % 10] %= mod;
                dp[i + 1][(j * a[i + 1]) % 10] += dp[i][j];
                dp[i + 1][(j * a[i + 1]) % 10] %= mod;
            }
        }
    }
    for (int i = 0; i < 10 ;i++)
        cout << dp[n][i] << "\n";
        return 0;
}