1233 - Coin Change (III)

1233 - Coin Change (III)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

In a strange shop there are n types of coins of valueA₁, A₂ ... A_n. C₁, C₂,... C_n denote the number of coins of value A₁, A₂... A_n respectively. You have to find the number of differentvalues (from 1 to m), which can be produced using these coins.

Input

Input starts with an integer T (≤ 20),denoting the number of test cases.

Each case starts with a line containing two integers n (1≤ n ≤ 100), m (0 ≤ m ≤ 10⁵). The nextline contains 2n integers, denoting A₁, A₂ ...A_n, C₁, C₂ ... C_n (1 ≤ A_i≤ 10⁵, 1 ≤ C_i ≤ 1000). All A_iwill be distinct.

Output

For each case, print the case number and the result.

Sample Input

2

3 10

1 2 4 2 1 1

2 5

1 4 2 1

Output for Sample Input

Case 1: 8

Case 2: 4

题意: 给出n中硬币的币值,每种硬币各有c[i]个,问1~m中有多少个能够由这n种硬币构成

题解: 这道题可以用多重背包做,也可以用完全背包做,不过要加一个计数器,保证取得个数不超过c[i]个,dp[i]表示能不能组成i,b[i]表示组成i的用了多少个a[i];

AC代码:

/* ***********************************************
Author        :xdlove
Created Time  :2015年11月04日 星期三 12时12分04秒
File Name     :DP/1233_Coin_Change_III.cpp
 ************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;


#define REP_ab(i,a,b) for(int i = a; i <= b; i++)
#define REP(i, n) for(int i = 0; i < n; i++)
#define REP_1(i,n) for(int i = 1; i <= n; i++)
#define DEP(i,n) for(int i = n - 1; i >= 0; i--)
#define DEP_N(i,n) for(int i = n; i >= 1; i--)
#define CPY(A,B) memcpy(A,B,sizeof(B))
#define MEM(A) memset(A,0,sizeof(A))
#define MEM_1(A) memset(A,-1,sizeof(A))
#define MEM_INF(A) memset(A,0x3f,sizeof(A))
#define MEM_INFLL(A) memset(A,0x3f3f,sizeof(A))
#define mid (((l + r) >> 1))
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1
#define ls (u << 1)
#define rs (u << 1 | 1)


typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 5;
const int MAXM = MAXN;
const int mod = 1e9 + 7;

int a[110],c[110],b[MAXN];
bool dp[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T,n,m,cnt = 0;
    cin>>T;
    while(T--)
    {
        scanf("%d %d",&n,&m);
        REP(i,n) scanf("%d",&a[i]);
        REP(i,n) scanf("%d",&c[i]);
        MEM(dp);
        dp[0] = true;
        REP(o,n)
        {
            MEM(b);
            REP_ab(i,a[o],m) 
            {
                if(dp[i] || b[i - a[o]] >= c[o]) continue;
                dp[i] |= dp[i - a[o]];
                b[i] = b[i - a[o]] + 1;
            }
        }
        int num = 0;
        REP_1(i,m)
            num += dp[i];
        printf("Case %d: %d\n",++cnt,num);
    }
    return 0;
}