ACM学习历程—CodeForces 176B Word Cut（字符串匹配 && dp && 递推）

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 word w = 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 exactly ksplit 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.

Input

The first line contains a non-empty word start, the second line contains a non-empty word end. The words consist of lowercase Latin letters. The number of letters in word start equals the number of letters in word end and is at least 2 and doesn't exceed 1000 letters.

The third line contains integer k (0 ≤ k ≤ 10⁵) — the required number of operations.

Output

Print a single number — the answer to the problem. As this number can be rather large, print it modulo 1000000007(10⁹ + 7).

Sample Input

 

Input

ab
ab
2

Output

1

Input

ababab
ababab
1

Output

2

Input

ab
ba
2

Output

0

Sample Output

 

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

题目大意就是问有几种操作方法，能在k次操作下得到目标串。

首先要肯定的是题目给的操作相当于一个循环串的位移。（位移不为0）

这样的话，就可以对目标串的第一位分析了。

记s(now, k)表示经过k次操作后，目标串的第一位在当前串的第now位。（初始k == 0时，由于串是循环串，会出现不止一个now的值不为0）

于是s(now, k) = sum{s(i, k-1)} (i != now)

只要不是本身就在now，都能位移到now。

大一时做这个题，到就只能想到这里，然后进行了O(k*strlen(str)*strlen(str))的暴力运算。

但是上述式子可以进行一定的变形：

s(now, k) = sum - s(now, k-1)

这样的话，只需要记录下sum，就能在O(k*strlen(str))的时间复杂度下完成了。

需要注意的是，由于采用一维dp会导致计算s(now)的时候破坏了前一个状态。

所以此处需要同时保存前一个状态和当前状态。所以开了第二维，且第二维仅保留两次的状态。（此处采用了亦或运算进行优化）

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#define N 1000000007

using namespace std;

char from[1005], to[1005];
int n, len, s[1005][2], sum;

bool Pair(int now)
{
    int i = 0;
    for (;;)
    {
        if (i == len)
            return true;
        if (from[now] != to[i])
            return false;
        i++;
        now = (now+1) % len;
    }
}

bool Input()
{
    if (scanf("%s", from) == EOF)
        return false;
    scanf("%s%d", to, &n);
    memset(s, 0, sizeof(s));
    len = strlen(from);
    sum = 0;
    for (int i = 0; i < len; ++i)
    {
        if (Pair(i))
        {
            s[i][0] = 1;
            sum++;
        }
    }
    return true;
}

void Work()
{
    int tem, state = 0;
    for (int i = 1; i <= n; i++)
    {
        tem = 0;
        state = state^1;
        for (int j = 0; j < len; ++j)
        {
            s[j][state] = sum - s[j][state^1];
            s[j][state] = (s[j][state]%N+N)%N;
            tem += s[j][state];
            tem %= N;
        }
        sum = tem;
    }
    printf("%d
", s[0][state]);
}

int main()
{
    //freopen("test.in", "r", stdin);
    while (Input())
    {
        Work();
    }
    return 0;
}