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题意:给你一个字符串,然后问你添加几个字符能凑成循环串。
思路:
很显然,k = m / ( m - Next[m]) 为已经出现的循环节次数,l = m - Next[m] 表示循环节长度。用总的长度m - l *k为某一循环节剩余长度,然后再用 l - 剩余长度就是答案。
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 Next[M+5];
int dp[M+5];
void kmp_pre(int m)
{
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()
{
int T;
read(T);
while(T--){
scanf("%s",s);
int m = strlen(s);
kmp_pre(m);
if(m%(m - Next[m]) == 0){
int k = m/(m - Next[m]);
if(k == 1) cout<<m<<endl;
else puts("0");
}
else {
int l = m - Next[m];
cout<<l - m%l<<endl;
}
}
}