UESTC 491 Tricks in Bits
Time Limit: 1000MS | Memory Limit: 65535KB | 64bit IO Format: %lld & %llu |
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Description
Given N unsigned 64-bit integers, you can bitwise NOT each or not. Then you need to add operations selected from bitwise XOR, bitwise ORand bitwise AND, between any two successive integers and calculate the result. Your job is to make the result as small as possible.
Input
The first line of the input is T (no more than 1000), which stands for the number of test cases you need to solve.
Then T blocks follow. The first line of each block contains a single number N (1≤N≤100) indicating the number of unsigned 64-bit integers. Then n integers follow in the next line.
Output
For every test case, you should output Case #k: first, where k indicates the case number and counts from 1. Then output the answer.
Sample Input
2 3 1 2 3 2 3 6
Sample Output
Case #1: 0
Case #2: 1
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
using namespace std;
typedef unsigned long long int LL;
int n;
LL ans;
LL MAX;
LL a[105];
LL min(LL a,LL b){return (a<b?a:b);}
void dfs(LL num,int cnt)
{
if(ans==0)
return;
if(num==0)
{
ans=0;
return;
}
if(cnt==n+1)
{
ans=min(ans,num);
return;
}
dfs(num|(~a[cnt]),cnt+1);
dfs(num&(~a[cnt]),cnt+1);
dfs(num^(~a[cnt]),cnt+1);
dfs(num|a[cnt],cnt+1);
dfs(num&a[cnt],cnt+1);
dfs(num^a[cnt],cnt+1);
}
int main()
{
int t;
scanf("%d",&t);
int cas=0;
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%llu",&a[i]);
MAX=1;
MAX<<=63;
ans=MAX;
dfs(a[1],2);
dfs(~a[1],2);
printf("Case #%d: %llu\n",++cas,ans);
}
return 0;
}