本文主要是介绍HDU3746-KMP循环节，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：题目链接
题意：给你一个字符串，要求将字符串的全部字符最少循环2次需要添加的字符数。
例子：
abcabc 已经循环2次，添加数为0
abcac 没有循环2次，添加字符abcac。数目为5.
abcabcab 已经循环过2次，但第三次不完整，需要添加数为1
next函数求法：
void getnext(const char *s)
{int i = 0, j = -1;nextval[0] = -1;while(i != len){if(j == -1 || s[i] == s[j])nextval[++i] = ++j;elsej = nextval[j];}
}
void getnext(const char *p) //前缀函数(滑步函数)
{int i = 0, j = -1;nextval[0] = -1;while(i != len){if(j == -1 || p[i] == p[j]) //(全部不相等从新匹配 || 相等继续下次匹配){++i, ++j;if(p[i] != p[j]) //abcdabcenextval[i] = j;else //abcabcanextval[i] = nextval[j];}elsej = nextval[j]; //子串移动到第nextval[j]个字符和主串相应字符比较}
}
就是求出匹配的串之后，对KMP的循环节和原来的长度比较：
如何利用KMP的next求字符串的循环节
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int gcd(int n,int m)
{if(n<m) swap(n,m);return n%m==0?m:gcd(m,n%m);
}
int lcm(int n,int m)
{if(n<m) swap(n,m);return n/gcd(n,m)*m;
}
#define N 100010
int prime[N];
struct node
{int x, y;
};
bool cmp(const node & a, const node & b)
{return a.x > b.x;
}
void getPrime();
void bash();
void wzf();
void SG();
int QuickMod(int a, int b, int n);char num[N];
int next[N];
int len;
void getnext()
{int i = 0, j = -1;next[0] = -1;while(i != len){if(j == -1 || num[i] == num[j])next[++i] = ++j;elsej = next[j];}
}int main()
{int t;scanf("%d", &t);while(t--){scanf("%s", num);len = strlen(num);getnext();int L = len - next[len];if(len != L && len%L == 0)printf("0\n");else{int ans = L - next[len]%L;printf("%d\n", ans);}}return 0;
}int QuickMod(int a,int b,int n)
{int r = 1;while(b){if(b&1)r = (r*a)%n;a = (a*a)%n;b >>= 1;}return r;
}void getPrime()
{memset(prime, 0, sizeof(prime));prime[0] = 1;prime[1] = 1;for(int i = 2; i < N; ++i){if(prime[i] == 0){for(int j = i+i; j < N; j+=i)prime[j] = 1;}}
}void bash(int n, int m)
{if(n%(m+1) != 0)printf("1\n");elseprintf("0\n");
}void wzf(int n, int m)
{if(n > m)swap(n, m);int k = m-n;int a = (k * (1.0 + sqrt(5.0))/2.0);if(a == n)printf("0\n");elseprintf("1\n");
}int numsg[N];
void SG(int n)
{int sum = 0;for(int i=0; i < n; i++){scanf("%d",&numsg[i]);sum ^= numsg[i];}if(sum == 0)printf("No\n");else{printf("Yes\n");for(int i = 0; i < n; i++){int s = sum ^ numsg[i];if(s < numsg[i])printf("%d %d\n", numsg[i], s);//从有num[i]个石子的堆后剩余s个石子}}
}
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