lightoj 1147 - Tug of War

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight
of the people on each team should be as nearly equal as possible.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

The first line of each case is a blank line. The next line of input contains an integer n (2 ≤ n ≤ 100), the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight
of person 2; and so on. Each weight is an integer between 1 and 100000. The summation of all the weights of the people in a case will not exceed 100000.

Output

For each case, print the case number and the total number weights of the people in two teams. If the weights differ, print the smaller weight first.

Sample Input	Output for Sample Input
2   3 100 90 200   4 10 15 17 20	Case 1: 190 200 Case 2: 30 32

题意是给你一些人的重量，让你把他们分成人数差最大为1的两组，要求保证两组重量的差最小，先输出小的那个。

dp[i] = x，x是一个二进制数，第k位是1的话表示有k个人的重量为i。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;

int main(void)
{
int T,n,i,j;
int a[110];
LL dp[100010];
scanf("%d",&T);
int cas = 1;
while(T--)
{
scanf("%d",&n);
int sum = 0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum += a[i];
}
memset(dp,0,sizeof(dp));
int V = sum/2;
dp[0] = 1;
for(i=1;i<=n;i++)
{
for(j=V;j>=a[i];j--)
{
dp[j] |= (dp[j-a[i]]<<1);
}
}
for(i=V;i>=0;i--)
{
if(dp[i]&(1LL<<(n+1)/2)||dp[i]&(1LL<<(n/2)))
{
printf("Case %d: %d %d\n",cas++,i,sum-i);
break;
}
}
}
return 0;
}