题目:
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 nn chips today, the ii-th chip produced this day has a serial number s_isi.
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:
\displaystyle max_{i, j, k}(s_i + s_j) \oplus s_kmaxi,j,k(si+sj)⊕sk
which i,j,ki,j,k are three different integers between 11 and nn. And \oplus⊕ is symbol of bitwise XOR.
Can you help John calculate the checksum number of today?
Input Format
The first line of input contains an integer TT indicating the total number of test cases.
The first line of each test case is an integer nn, indicating the number of chips produced today.
The next line has nn integers s_1,s_2,..,s_ns1,s2,..,sn, separated with single space, indicating serial number of each chip.
• 1\le T \le 10001≤T≤1000
• 3\le n\le 10003≤n≤1000
• 0\le s_i \le 10^90≤si≤109
• There are at most 1010 testcases with n > 100n>100
Output Format
For each test case, please output an integer indicating the checksum number in a line.
样例输入复制
2 3 1 2 3 3 100 200 300
样例输出复制
6 400
分析:
看到这道题的通过率这么高,我就想用简单的方法————暴力试一下,结果就过了;
代码:
#include<stdio.h> #include<math.h> #include<algorithm> using namespace std; int a[1005]; int main() { int t,n,i,j,k,maxx; scanf("%d",&t); while(t--) { maxx=0; scanf("%d",&n); for(i=1;i<=n;i++) scanf("%d",&a[i]); for(k=1;k<=n;k++) for(i=k+1;i<=n;i++) for(j=i+1;j<=n;j++) { maxx=max(maxx,(a[k]+a[j])^a[i]); maxx=max(maxx,(a[i]+a[j])^a[k]); maxx=max(maxx,(a[i]+a[k])^a[j]); } printf("%d\n",maxx); } return 0; }