Codeforces Round #607 (Div. 2) C. Cut and Paste

Posted 2021-03-20 ydddd

链接：

https://codeforces.com/contest/1281/problem/C

题意：

outputstandard output
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by si.

There is one cursor. The cursor‘s location ? is denoted by an integer in {0,…,|s|}, with the following meaning:

If ?=0, then the cursor is located before the first character of s.
If ?=|s|, then the cursor is located right after the last character of s.
If 0<?<|s|, then the cursor is located between s? and s?+1.
We denote by sleft the string to the left of the cursor and sright the string to the right of the cursor.

We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:

The Move action. Move the cursor one step to the right. This increments ? once.
The Cut action. Set c←sright, then set s←sleft.
The Paste action. Append the value of c to the end of the string s. Note that this doesn‘t modify c.
The cursor initially starts at ?=0. Then, we perform the following procedure:

Perform the Move action once.
Perform the Cut action once.
Perform the Paste action s? times.
If ?=x, stop. Otherwise, return to step 1.
You‘re given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 109+7.

It is guaranteed that ?≤|s| at any time.

思路：

模拟操作到x的长度，剩下的遍历一遍算值，居然不超时。。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int x;
        string s;
        cin >> x >> s;
        int l = 1;
        while((int)s.size() < x)
        {
            int len = s.size();
            for (int i = 1;i < (int)s[l-1]-'0';i++)
                s += s.substr(l, len-l);
            l++;
        }
        LL ans = s.size()%MOD;
        for (int i = l;i <= x;i++)
        {
            int tmp = s[i-1]-'0';
            ans = (ans+(ans-i)*(tmp-1)%MOD+MOD)%MOD;
        }
        cout << ans%MOD << endl;
    }

    return 0;
}