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  • HDU 5536--Chip Factory(暴力)

    Chip Factory

    Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 5394    Accepted Submission(s): 2422


    Problem Description
    John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, the i-th chip produced this day has a serial number si.
    At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:
    maxi,j,k(si+sj)sk
    which i,j,k are three different integers between 1 and n. And  is symbol of bitwise XOR.
    Can you help John calculate the checksum number of today?
     
    Input
    The first line of input contains an integer T indicating the total number of test cases.

    The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1,s2,..,sn, separated with single space, indicating serial number of each chip.

    1T1000
    3n1000
    0si109
    There are at most 10 testcases with n>100
     
    Output
    For each test case, please output an integer indicating the checksum number in a line.
     
    Sample Input
    2 3 1 2 3 3 100 200 300
     
    Sample Output
    6 400
     
    Source
     
    时间充裕,暴力一发0.0
     
     1 #include <iostream>
     2 #include<cstring>
     3 #include<algorithm>
     4 #include<cstdio>
     5 #include<cmath>
     6 using namespace std;
     7 int a[1005];
     8 int main()
     9 {
    10     int t;
    11     scanf("%d",&t);
    12     while(t--)
    13     {
    14         int n,m=-999;
    15         scanf("%d",&n);
    16         for(int i=0;i<n;i++)
    17             scanf("%d",&a[i]);
    18         for(int i=0;i<n;i++){
    19             for(int j=i+1;j<n;j++){
    20                 for(int k=j+1;k<n;k++){
    21                     m=max(m,(a[i]+a[j])^a[k]);
    22                     m=max(m,(a[j]+a[k])^a[i]);
    23                     m=max(m,(a[i]+a[k])^a[j]);
    24                 }
    25             }
    26         }
    27         printf("%d
    ",m);
    28     }
    29     return 0;
    30 }
    View Code
     
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  • 原文地址:https://www.cnblogs.com/FlyerBird/p/9740342.html
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