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1108: Problem D: Tug of War

发表于 2017-10-06   |   分类于 HUSTOJ   |   阅读次数 1,279

题目描述

Problem D: 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.
The first line of input contains n 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 450. There are at most 100 people at the picnic.
Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

输入

暂无

输出

暂无

样例输入

3
100
90
200

样例输出

190 200

参考代码

#include <stdio.h>
#include <stdlib.h>
int people[1000];
int i,j,k,n,tot,best;
char cando[101][50000];
main() 
{
    scanf("%d",&n);
    for (i=0;i<n;i++) 
    {
        scanf("%d",&people[i]);
        tot += people[i];
    }
    cando[0][0] = 1;
    for (i=0;i<n;i++) 
    {
        /* try every person */
        for (j=n/2;j>=0;j--) 
        {
            for (k=45000;k>=0;k--) 
            {
                if (cando[j][k]) cando[j+1][k+people[i]]=1;
            }
        }
    }
    for (i=0;i<=45000;i++) 
    {
        if (!cando[n/2][i]) continue;
        if (abs(tot-2*i) < abs(tot-2*best)) best = i;
    }
    if (best > tot-best) best = tot-best;
    printf("%d %dn",best,tot-best);
}

解析

暂无

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若是凉夜已成梦

若是凉夜已成梦

青春里 总有些事情要努力去做 总有些梦想要拼命去追。

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