hdu2126（三维背包）

思路：显然的思路是前i个物品选j个花费k元的次数。

dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - v[i]];

由上面的转移方程可以看出来，可以去掉一维，就是dp[j][k] = dp[j - 1][k - v[i]] + dp[j][k];

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
inline int Readint(){
	char c = getchar();
	while(!isdigit(c)) c = getchar();
	int x = 0;
	while(isdigit(c)){
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x;
}
int n,m,t;
int a[35], dp[35][510];
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		memset(dp, 0,sizeof dp);
		for (int i = 1;i <= n;i++)
			scanf("%d",&a[i]);
		dp[0][0] = 1;
		for (int i = 1;i <= n;i++){
			for (int j = n;j >= 1;--j){
				for (int k = m;k >= a[i];--k)
					dp[j][k] += dp[j - 1][k - a[i]];
			}
		}
		int ans;
		bool flag=false;
		for (int j = n;j >= 1;--j){
			ans = 0;
			for (int k = m;k >= 0;k--)
				ans += dp[j][k];
			if (ans){
				printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", ans, j);
				flag = true;
				break;
			}
		}
		if (!flag) printf("Sorry, you can't buy anything.\n");
	}
	return 0;
}