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  • Hdu 1358 Period

    Period

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 13128    Accepted Submission(s): 6133

    Problem Description

    For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

    Input

    The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

    Output

    For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

    Sample Input

    3 aaa

    12 aabaabaabaab

    0

    Sample Output

    Test case #1

    2 2

    3 3

    Test case #2

    2 2

    6 2

    9 3

    12 4

    #include <bits/stdc++.h>
    #define K 1000000+9
    int nt[1000000+9];
    char a[K];
    void kmp_next(char *T,int *nt)
    {
        nt[1]=0;
        for(int i=2,j=0,m=strlen(T);i<=m;i++)
        {
            while(j&&T[i-1]!=T[j])j=nt[j-1+1];
            if(T[i-1]==T[j])j++;
            nt[i]=j;
        }
    }
    int main()
    {
        int icase=1;
        int N;
        while(scanf("%d",&N)==1)
        {
        if(N==0)break;
        printf("Test case #%d
    ",icase++);
        scanf("%s",a);
        kmp_next(a,nt);
        for(int i=2;i<=N;i++)
        {
            if(i%(i-nt[i])==0&&nt[i]!=0)printf("%d %d
    ",i,i/(i-nt[i]));
        }
        printf("
    ");
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/linruier/p/9726789.html
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