CodeForces-1281C Cut and Paste

传送门：CodeForces-1281C

思路分析

实际上，每次复制的次数都只跟前面n个字符有关，所以只要构造出长度为n的字符串，之后的长度都可以通过计算得到

样例输入

4
5
231
7
2323
6
333
24
133321333

样例输出

25
1438
1101
686531475

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define vi vector<int>
#define lowbit(x) x&(-x)
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

using namespace std;

const int INF = 0x3f3f3f3f;
const int mod=1e9+7;

int main() {
    IOS;
#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    cin>>t;
    while(t--) {
        int n;
        string s;
        cin>>n>>s;
        int l=1;
        int len=s.size();
        while(len<n) {
            for(int i=1; i<s[l-1]-'0'; i++)
                s+=s.substr(l,len-l);
            len=s.size();
            l++;
        }

        ll ans=s.size()%mod;
        for(int i=l; i<=n; i++)
            ans=(ans+1LL*(s[i-1]-'0'-1)%mod*(ans-i)%mod+mod)%mod;
        cout<<ans<<endl;
    }




    return 0;
}