Uva1328（kmp）

题目链接

 

讲解引用博客

 循环节部分有待继续理解

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
char s[maxn];
int f[maxn];


void solve()
{
    int m = strlen(s);
    f[0] = f[1] = 0;
    for(int i = 1; i < m; i++)
    {
        int j = f[i];
        while(j && s[i] != s[j])    j = f[j];
        f[i + 1] = s[i] == s[j] ? j + 1 : 0;    //例如：abcdabe,e失配的话会回溯到c,因为有公共前缀ab
    }
}


int main()
{
    cin >> s;
    solve();
    for(int i = 0; i < strlen(s); i++)
        cout << i << " " << f[i] << endl;
    int n = strlen(s);
    /*
    for(int i = 0; i <= n; i++)
        cout << f[i] << "  ";
    cout << endl;
    */

    for(int i = 2; i <= n; i++)
    {
        if(f[i] > 0 && i % (i - f[i]) == 0)    
            printf("%d %d\n", i, i / (i - f[i]));
        //if(f[i] > 0 && (f[i] + 1) % (i - f[i]) == 0)
            //printf("%d %d\n", i + 1, (f[i] + 1) / (i - f[i]) + 1);    个人理解也可以
    }
    return 0;
}