POJ 1961 Period (KMP)

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解题思路：

利用next 数组的性质求解重复子串。循环节的长度为i - next[i];

#include <iostream>#include <cstring>#include <cstdlib>#include <vector>#include <cmath>#include <algorithm>#include <cstdio>using namespace std;const int maxn = 1000000 + 10;char s[maxn];int n;int next[maxn];int main(){    int kcase = 1;    while(scanf("%d", &n)!=EOF)    {        if(n == 0) break;        scanf("%s", s);        next[0] = 0; next[1] = 0;        for(int i=1;i<n;i++)        {            int j = next[i];            while(j && s[i] != s[j]) j = next[j];            next[i+1] = (s[i] == s[j]) ? j + 1 : 0;        }        printf("Test case #%d\n", kcase++);        for(int i=2;i<=n;i++)        {            if(next[i] > 0 && i % (i - next[i]) == 0)            {                printf("%d %d\n", i, i / (i - next[i]));            }        }        printf("\n");    }    return 0;}

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