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题意:给你 a = "abc" and b = "def" then a*b = "abcdef",定义字符串相乘,再给你 a^0 = "" (the empty string) and a^(n+1) = a*(a^n). 定义字符串的幂。然后给你一个字符串,求n最大是多少。
思路:n最大是多少,就是说最多是多少个循环节所构成这个字符串。所以求完Next数组直接判断即可。
AC Code:
#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<cstdio>
#include<iomanip>
#include<sstream>
#include<algorithm>
using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define sRead(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define gc(x) scanf(" %c",&x);
#define mmt(x,y) memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define INF 0x3f3f3f3f
#define ll long long
#define mod 998244353
#define pdd pair<double,double>
const int N = 1000;
const int M= 1e6;
char s[M+5];
int m;
int Next[M+5];
void kmp_pre()
{
int i = 0;
int j = Next[0] = -1;
while(i < m){
while(j != -1&&s[i]!=s[j]) j = Next[j];
Next[++i] = ++j;
}
}
int main()
{
while(~scanf("%s",s)&&s[0]!='.'){
mmt(Next,0);
m = strlen(s);
kmp_pre();
if(m % (m - Next[m]) == 0) cout<<m/(m - Next[m])<<endl;
else puts("1");
}
}