HZNU Training 3—M"/>
HZNU Training 3—M
M - Word Cut
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 andy 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 k split 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.
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 least2 and doesn't exceed 1000 letters.
The third line contains integer k (0 ≤ k ≤ 105) — the required number of operations.
Print a single number — the answer to the problem. As this number can be rather large, print it modulo 1000000007 (109 + 7).
ab ab 2Output
1Input
ababab ababab 1Output
2Input
ab ba 2Output
0
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
f[i+1][1]=(((len-x)*f[i][0])%MOD+((len-x-1)*f[i][1])%MOD)%MOD;
【代码】
#include <iostream>
#include <cstdio>
#include <cstring>
#define MOD 1000000007
using namespace std;
long long f[100000][2];
char a[1000],b[1000];
int main()
{ gets(a);gets(b);int len=strlen(a);if (strncmp(a,b,len)==0) f[0][0]=1;else f[0][1]=1; int x=0; for (int i=0;i<len;i++) { for (int j=0;j<len;j++) if (a[(i+j)%len]!=b[j]) goto out; x++;out:; } int n;scanf("%d",&n);for (int i=0;i<n;i++) { f[i+1][0]=(x*f[i][1]+(x-1)*f[i][0])%MOD; f[i+1][1]=(((len-x)*f[i][0])%MOD+((len-x-1)*f[i][1])%MOD)%MOD; } printf("%d\n",f[n][0]);
}
#include <iostream>
#include <cstdio>
#include <cstring>
#define MOD 1000000007
using namespace std;
long long f[100000][2];
char a[1000],b[1000];
int main()
{ gets(a);gets(b);int len=strlen(a);if (strncmp(a,b,len)==0) f[0][0]=1;else f[0][1]=1; int x=0; for (int i=0;i<len;i++) { for (int j=0;j<len;j++) if (a[(i+j)%len]!=b[j]) goto out; x++;out:; } int n;scanf("%d",&n);for (int i=0;i<n;i++) { f[i+1][0]=(x*f[i][1]+(x-1)*f[i][0])%MOD; f[i+1][1]=(((len-x)*f[i][0])%MOD+((len-x-1)*f[i][1])%MOD)%MOD; } printf("%d\n",f[n][0]);
} 更多推荐
HZNU Training 3—M
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