题意：给你a串和b串，你能切k次，每次切完将尾部分放在头的前面，问有多少种方案切k次从a串变为b串
思路：令dp[i][0]为砍了i次变成b串的方案数，dp[i][1]为砍了i次变成非b串的方案数，然后预处理一下前缀就可以DP了
#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
#define LL long long
LL dp[300000][2];
int main()
{string a,b;int k;cin >> a >> b >>k;int lena = a.size();int lenb = b.size();if (a==b)dp[0][0]=1;else dp[0][1]=1;int x = 0;for (int i = 0;i<lena;i++){int flag = 1;for (int j = 0;j<lena;j++){if (a[(i+j)%lena] != b[j]){flag = 0;break;}}if (flag)x++;}for (int i = 0;i<k;i++){dp[i+1][0]=(x*dp[i][1]+(x-1)*dp[i][0])%mod;dp[i+1][1]=((lena-x)*dp[i][0]+(lena-x-1)*dp[i][1])%mod;}printf("%d\n",dp[k][0]);
}
Description
Let's consider one interesting word game. In this game you should transform one word into another through special operations.
Let's say we have word w, let's split this word into two non-empty parts x and y so, that w = xy. A split operation is transforming wordw = xy into word u = yx. For example, a split operation can transform word "wordcut" into word "cutword".
You are given two words start and end. Count in how many ways we can transform word start into word end, if we apply exactlyksplit operations consecutively to word start.
Two ways are considered different if the sequences of applied operations differ. Two operation sequences are different if exists such number i (1 ≤ i ≤ k), that in the i-th operation of the first sequence the word splits into parts x and y, in the i-th operation of the second sequence the word splits into parts a and b, and additionally x ≠ a holds.
Output
Print a single number — the answer to the problem. As this number can be rather large, print it modulo 1000000007(109 + 7).
Hint
The sought way in the first sample is:
ab → a|b → ba → b|a → ab
In the second sample the two sought ways are:
- ababab → abab|ab → ababab
- ababab → ab|abab → ababab
)
