HDU 5536 Chip Factory(暴力+优化)
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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
whichi,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?
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:
which
Can you help John calculate the checksum number of today?
The first line of each test case is an integer
There are at most
231 2 33100 200 300
6400
题解:
题意:
给你一个序列,找出三个不同的i,j,k,使得(a[i]+a[j])^a[k]最大
思路:
暴力+一些剪枝可以过
代码:
#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>#define ll long long#define INF 100861111#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1using namespace std;int a[1005];int main(){ int i,j,k,d,test,maxx,n; scanf("%d",&test); while(test--) { scanf("%d",&n); maxx=0; for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) { for(j=i+1;j<n;j++) { for(k=j+1;k<n;k++) { maxx=max(maxx,(a[i]+a[j])^a[k]); maxx=max(maxx,(a[j]+a[k])^a[i]); maxx=max(maxx,(a[i]+a[k])^a[j]); } } } printf("%d\n",maxx); }}
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