HDU5562——Clarke and food_hdu clarke and food-优快云博客

本文介绍了一道关于背包问题的经典算法题目，通过购物场景描述了如何利用贪心算法选择尽可能多的食物放入有限容量的背包中。该问题适用于初学者理解和实践贪心算法的基本思想。

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Clarke and food

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1002 Accepted Submission(s): 563

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a cook, was shopping for food.
Clarke has bought

n food. The volume of the

ith food is

vi. Now Clarke has a pack with volume

V. He wants to carry food as much as possible. Tell him the maxmium number he can brought with this pack.

Input

The first line contains an integer

T(1≤T≤10), the number of the test cases.
For each test case:
The first line contains two integers

n,V(1≤n≤105,1≤V≤109).
The second line contains

n integers, the

ith integer denotes

vi(1≤vi≤109).

Output

For each test case, print a line with an integer which denotes the answer.

Sample Input

1
3 5
1 3 4

Sample Output

2

Hint:  
We can carry 1 and 3, the total volume of them is 5.

解：题意为有n个食物体积分别为a[i],要将他们装入体积为v的书包内，问能够最多装多少个食物，优先装小的食物。贪心水题。

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;

int main()
{
	int t;
	long long n,v;
	long long a[100005];
	scanf("%d",&t);
	while(t--){
		int count=0;
		scanf("%lld%lld",&n,&v);
		for(int i=0;i<n;i++){
			scanf("%lld",&a[i]);
		}
		sort(a,a+n);
		for(int i=0;i<n;i++){
			if(v-a[i]>=0){
				v=v-a[i];
				count++;
			}
			else
				break;
		}
		printf("%d\n",count);
	}
	
	return 0;
}