宁波工程学院[1372] Do What n个数中取出某些数使得和大于T且和最小

[1372] Do What

时间限制: 1000 ms　内存限制: 65535 K

问题描述

There are n numbers of business, different business will cost you different times. But you need not to finish all of them, just it is ok when you finish some of them that you spend more than T minutes(do not include T minutes).
So, can you find the way to cost the minimum time when pass T times?

输入

Input until EOF.
First line will contain a positive integer n(1 <= n <= 100) means the number of businesses.
Next line follows n positive integers ti(1 <= ti <= 10), means the time of each business you will cost.
Last line is a integer T(1 <= T <= 1000).

输出

The minimum time of you cost, if there is no answer, print '-1'.

样例输入

3
1 2 3
10
3
1 4 7
6

样例输出

-1
7

提示

无

来源

Hungar

操作

 

 

题意： 从n个数中取出任意个 数字 使得这些数字的和大于T且和最小

 

思路：

01背包  背包过程中取离T最近的值

 

详细看代码

#include<stdio.h>
#include<string.h>
int n,t,ans;
int val[105];
int dp[10000];
void _01bag(int v)
{
	int i,j;
	memset(dp,0,sizeof(dp));
	for(i=0;i<n;i++)
		for(j=v;j>=val[i];j--)
			if(dp[j]<dp[j-val[i]]+val[i])
			{
				dp[j]=dp[j-val[i]]+val[i];
				if(dp[j]>t&&ans>dp[j]-t) ans=dp[j]-t;
			}
			printf("%d\n",ans+t);
}
int main()
{
	int i;
	while(scanf("%d",&n)!=EOF)
	{
		ans=999999999;
		int sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&val[i]);
			sum+=val[i];
		}
        scanf("%d",&t);
		if(sum>t)
		   _01bag(sum);
		else printf("-1\n");
	}
	return 0;
}