【无标题】

#include<iostream>
using namespace std;
struct inform{
    int num;
    string name;
}arr[100010];
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++){
        cin>>arr[i].num>>arr[i].name;
    }
    int t=0;
    while(m--){
        int a,b;
        cin>>a>>b;
        if((arr[t].num==0&&a==0)||(arr[t].num==1&&a==1))
            t=(t+n-b)%n;
        else
            t=(t+n+b)%n;
    }
    cout<<arr[t].name;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int ans=0;
    for(int i=0;i<=n/5;i++)
    {
        for(int j=0;j<=n/2;j++)
        {
            for(int l=0;l<=n;l++)
            {
                if(5*i+2*j+l==n)
                    ans++;
                else if(5*i+2*j+l>n)
                    break;
            }
        }
    }
    cout<<ans;
}

#include<iostream>
using namespace std;
const int N = 210;
int f[N];
int main()
{
    int w[] = {1, 2, 5};
    int n; cin >> n;
    f[0] = 1;
    for(int i = 0; i < 3; i ++ )
        for(int j = w[i]; j <= n; j ++ )
            f[j] += f[j - w[i]];//先累加1的，再累加2的，再累加5的
    cout << f[n] << endl;
    return 0;
}