period题解（用KMP算法来（判断字符串重复）

总Time Limit:

3000ms

Memory Limit:

65536kB

Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A ^K,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

#include

char th[1000002];
int next[1000001];
void makenext(char th[], int n)
{
    int i,j;
    next[1]=0;
    for(i=2;i<=n+1;i++)
    {
        int j=i-1;
        while(next[j]!=0&&th[next[j]-1]!=th[i-2]){
            j=next[j];
        }
        if(next[j]==0)
            next[i]=1;
        else next[i]=next[j]+1;
    }
}
int main()
{
    int n,i,j,N=1;
    while(scanf("%d",&n)!=EOF&&n){
            if(N!=1)
            printf("\n");
            getchar();
            gets(th);
            makenext(th,n);
        printf("Test case #%d\n",N);
        N++;
        for(i=2;i<=n;i++){
            if(i%(i-next[i+1]+1)==0&&i/(i-next[i+1]+1)>1){
                printf("%d %d\n",i,i/(i-next[i+1]+1));
            }
        }
    }
    return 0;

}

next数组有两种，一种是以-1开头，另一种是以0开头，以0开头的和-1的写法有点不一样，但理解了意思就不难写出来，以-1开头的next数组中存的是字符串的下标，但以0开头的next数组存的是字符串的位数，求到第i位字符的最小循环节，有公式：i-next[i]+1(next数组以0开头)，以-1开头的是i-next[i],注意这个循环节不一定是严格意义上的循环节，如果i%(i-next[i]+1)==0,表明到第i个字符为止，可以有循环节整数倍重复的到，重复次数为i/(i-next[i]+1)