ACM: 限时训练题解-Heavy Coins-枚举子集-暴力枚举

1. Heavy Coins

 

Bahosain has a lot of coins in his pocket. These coins are really heavy, so he always tries to get rid of some of the coins by using them when paying for the  taxi.

 

Whenever Bahosain has to pay S pennies for the taxi driver, he tries to choose the maximum number of coin pieces to pay. The driver will accept receiving more than S pennies only if he can’t remove one or more of the given coins and still has S or more  pennies.

For example, if Bahosain uses the coins of the following values: 2, 7 and 5 to pay 11 pennies, the taxi driver    will not accept this because the coin of value 2 can be removed. On the other hand, when Bahosain uses coins  of 7 and 5 to pay 11 pennies, the driver will accept  it.

 

Note that the driver won’t give Bahosain any change back if he receives more than S pennies, and Bahosain doesn’t care!

 

Input

 

The first line of input contains T (1 ≤ T ≤ 1001), the number of test   cases.

 

The first line of each test case contains two integers: N (1 ≤ N ≤ 10) and S (1 ≤ S ≤ 1000), where N is the number of coins in Bahosain’s pocket and S is the amount (in pennies) Bahosain has to pay for the taxi driver.

The next line contains N space-separated integers between 1 and 100 that represent the values (in pennies) of the coins in Bahosain’s  pocket.

 

Output

 

For each test case, print a single line with the maximum number of coins Bahosain can use to pay for the driver.

 

 

 

Sample Input						Sample Output
2						3
5	9					6
4	1   3	5	4
7	37
7	5   8	8	5	10	4

Note

In the first test case, Bahosain can pay in any of the following ways: (1, 3, 5), (3, 4, 4) or (1, 4, 4).

/*
题意：
A有一大堆的硬币，他觉得太重了，想花掉硬币去坐的士；
的士司机可以不找零，但是的士司机也不会多收零钱。
怎么样才能使 A 花的零钱最多。 
 
思路-暴力
2^10*1000=1024000  102W 完全不会超时。
直接暴力枚举 

方法： 
枚举子集，用二进制思想来枚举子集。 
 
*/ 

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define memset(a,b) memset(a,b,sizeof(a))
#define MX 10000 + 50
using namespace std;
int maxx,n,m,a[MX];

bool cmp(int x,int y) {
	return x>y;
}

void dfs(int x,int sum,int num) {
	if(sum>=m) {
		maxx=max(maxx,num);	//如果花掉的硬币值已经够坐车了，更新最大花掉的硬币数 
		return ;
	}
	if(x==n)return;  //如果已经查询了所有的硬币返回。 
	dfs(x+1,sum+a[x],num+1);  //递归下去 搜索取下一个硬币的情况 
	dfs(x+1,sum,num);        //搜索不取下一个硬币的情况
}

int main() {
	int T,num;
	cin>>T;
	while(T--) {
		scanf("%d%d",&n,&m);
		for(int i=0; i

　　

 

转载于:https://www.cnblogs.com/HDMaxfun/p/5709521.html