篇首语：本文由编程笔记#小编为大家整理，主要介绍了Codeforces Round #607 (Div. 2) C. Cut and Paste相关的知识，希望对你有一定的参考价值。

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
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
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() {
int len = s.size();
for (int i = 1;i <(int)s[l-1]-&＃39;0&＃39;;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]-&＃39;0&＃39;;
ans = (ans+(ans-i)*(tmp-1)%MOD+MOD)%MOD;
}
cout < }
return 0;
}