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  • HDU 5311 Sequence

    Hidden String

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 803    Accepted Submission(s): 302


    Problem Description
    Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n. He wants to find three nonoverlapping substrings s[l1..r1]s[l2..r2]s[l3..r3] that:

    1. 1l1r1<l2r2<l3r3n

    2. The concatenation of s[l1..r1]s[l2..r2]s[l3..r3] is "anniversary".
     

    Input
    There are multiple test cases. The first line of input contains an integer T (1T100), indicating the number of test cases. For each test case:

    There's a line containing a string s (1|s|100) consisting of lowercase English letters.
     

    Output
    For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).
     

    Sample Input
    2 annivddfdersewwefary nniversarya
     

    Sample Output
    YES NO
     

    Source
     

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    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<math.h>
    
    using namespace std;
    
    char str[10001];
    struct node
    {
        int x;
        char s;
    } q[10010];
    
    int main()
    {
        int T;
        char a[] = "anniversary";
        scanf("%d",&T);
        while(T--)
        {
            int flag = 0;
            scanf("%s",str);
            int l = strlen(str);
            int ll = strlen(a);
            int num = 0;
            for(int j=0; j<l; j++)
            {
                for(int i=1; i<=ll; i++)
                {
                    if(strncmp(str+j,a+num,i) == 0)
                    {
                        int num1 = num + i;
                        for(int jj=j+i; jj<l; jj++)
                        {
                            for(int ii=1; ii<=ll-i; ii++)
                            {
                                if(strncmp(str+jj,a+num1,ii) == 0)
                                {
                                    int num2 = num1 + ii;
                                    for(int jjj=jj+ii; jjj<l; jjj++)
                                    {
                                        if(strncmp(str+jjj,a+num2,ll-num2) == 0)
                                        {
                                            printf("YES
    ");
                                            flag = 1;
                                            break;
                                        }
                                    }
                                }
                                if(flag == 1)
                                {
                                    break;
                                }
                            }
                            if(flag == 1)
                            {
                                break;
                            }
                        }
                        if(flag == 1)
                        {
                            break;
                        }
                    }
                    if(flag == 1)
                    {
                        break;
                    }
                }
                if(flag == 1)
                {
                    break;
                }
    
            }
            if(flag == 0)
            {
                printf("NO
    ");
            }
        }
        return 0;
    }


     
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  • 原文地址:https://www.cnblogs.com/tlnshuju/p/7072773.html
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