POJ - 1961 最小循环节

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e6+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
int nxt[maxn];
char P[maxn];
void buildNext(){
    nxt[1]=0;
    int j=0,m=strlen(P+1);
    rep(i,2,m){
        while(j&&P[i]!=P[j+1])j=nxt[j];
        if(P[i]==P[j+1])j++;
        nxt[i]=j;
    } 
}
int main(){
    int m,kase=0;
    while(~iin(m)){
        if(m==0)break;
        s1(P);
        buildNext();
        printf("Test case #%d\n",++kase);
        rep(i,2,m){
            if(i%(i-nxt[i])==0&&i/(i-nxt[i])>1){
                print(i); blank;
                println((i/(i-nxt[i])));
            }
        }
        putchar(‘\n‘);
    }
    return 0;
}