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  • G

     
    Time Limit:4000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    Description

    Given a permutation P of 1 to N, YY wants to know whether there exists such three elements P[i1], P[i2], P[i3] that
    P[i1]-P[i2]=P[i2]-P[i3], 1<=i1<i2<i3<=N.
     

    Input

    The first line is T(T<=60), representing the total test cases.
    Each test case comes two lines, the former one is N, 3<=N<=10000, the latter is a permutation of 1 to N.
     

    Output

    For each test case, just output 'Y' if such i1, i2, i3 can be found, else 'N'.
     

    Sample Input

    2 3 1 3 2 4 3 2 4 1
     

    Sample Output

    N Y
     
     1 #include<cstdio>
     2 #include<string.h>
     3 using namespace std;
     4 int hush[10005];
     5 int str[10005];
     6 int main()
     7 {
     8     int t,n;
     9     while(scanf("%d",&t)!=EOF)
    10     {
    11         while(t--)
    12         {
    13             scanf("%d",&n);
    14             for(int i=1; i<=n; i++)
    15             {
    16                 scanf("%d",&str[i]);//把1——N保存到数组里
    17                 hush[str[i]]=i;//记录每个数的位置
    18             }
    19             int flag=0;
    20             for(int i=1; i<=n; i++)
    21             {
    22                  for(int j=i+1; j<=n; j++)
    23                 {
    24                    int temp=str[i]+str[j];//相加
    25                    if(temp%2)//优化,必须是偶数才行,尽量别写成temp%2==1,直接写temp%2
    26                      continue;
    27                    if(hush[temp/2]>i&&hush[temp/2]<j)
    28                    {
    29                        flag=1;
    30                        break;
    31                    }
    32                 }
    33                 if(flag==1) break;
    34             }
    35             printf("%c",flag==1?'Y':'N');
    36             printf("
    ");
    37         }
    38     }
    39     return 0;
    40 }
     
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  • 原文地址:https://www.cnblogs.com/angledamon/p/3894128.html
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