POJ 2406Power Strings(KMP)

POJ 2406

其实就是一个简单的kmp应用：

ans = n % (n - f[n]) == 0 ? n / (n - f[n]) : 1

其中f是失配函数

//#pragma comment(linker, "/STACK:1677721600")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf (-((LL)1<<40))
#define lson k<<1, L, (L + R)>>1
#define rson k<<1|1,  ((L + R)>>1) + 1, R
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define FIN freopen("in.txt", "r", stdin)
#define FOUT freopen("out.txt", "w", stdout)
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define dec(i, a, b) for(int i = a; i >= b; i --)

template<class T> T MAX(T a, T b) { return a > b ? a : b; }
template<class T> T MIN(T a, T b) { return a < b ? a : b; }
template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

//typedef __int64 LL;
typedef long long LL;
const int MAXN = 1000000 + 10;
const int MAXM = 110000;
const double eps = 1e-8;
LL MOD = 1000000007;

char s[MAXN];
int f[MAXN];

void get_next(int n) {
    mem0(f);
    rep (i, 1, n - 1) {
        int j = f[i];
        while(j && s[i] != s[j])  j = f[j];
        f[i + 1] = s[i] == s[j] ? j + 1 : 0;
    }
}

int main()
{
    //FIN;
    while(~scanf("%s", s) && s[0] != '.') {
        int n = strlen(s);
        get_next(n);
        printf("%d
", n % (n - f[n]) == 0 ? n / (n - f[n]) : 1);
    }
    return 0;
}