HDU3746-KMP循环节

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 100010int 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;        else        j = 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");    else        printf("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");    else        printf("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        }    }}