HDU 5562 Clarke and food(排序)——BestCoder Round #62(div.2)
Clarke and food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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 i th 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.
Clarke has bought n food. The volume of the i th 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 i th integer denotes vi(1≤vi≤109) .
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 i th 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.
Source
BestCoder Round #62 (div.2)
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附上该题对应的中文题
Clarke and food
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
问题描述
克拉克是一名人格分裂患者。某一天，克拉克变成了一个厨师，在采购食品。 克拉克一共采购了n个食物，每个食物有一个体积vi。现在克拉克有一个容量为V背包，他想用这个背包来装尽量多的食物。请你告诉他最多能装多少食物。
输入描述
第一行一个整数T(1≤T≤10)，表示数据的组数。 每组数据第一行是两个整数n,V(1≤n≤105,1≤V≤109)，表示克拉克的数目和背包容量。 接下来一行有n个正数，第i个整数表示vi(1≤vi≤109)。
输出描述
每组数据输出一行一个数，表示答案。
输入样例
1 3 5 1 3 4
输出样例
2
Hint
我们可以放第1个和第3个，这样总容量为5。
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出题人的解题思路：
Clarke and food
可以证明从小到大一直拿，拿到不能拿为止是最优的。
所以排序一下即可。
说实话，第一眼看到的时候觉得是01背包，然而，V的范围果断让我否定了这种想法
好吧，其实这题没有想象的那么难，要装尽可能多的食物，当然先装体积小的咯，所以我们先对食物的体积排序，然后依次放进背包就可以了
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 100005;
const int M = 100005;
const int inf = 100000000;
const int mod = 2009;
int s[N];
int main()
{
int t,n,V,i,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&V);
for(k=i=0;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
for(i=0;i<n;i++)
if(V-s[i]>=0)
{
k++;
V-=s[i];
}
else
break;
printf("%d\n",k);
}
return 0;
}
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