877

# 题目

## Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

## Input

The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

## Output

For each test case output the answer on a single line.

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

8
4

# 题解

**>多重背包问题<**

dp[i] 表示在钱数最多为 i 的情况下,能拼成的最大钱数

# 代码

/*
By:OhYee
Github:OhYee
HomePage:http://www.oyohyee.com
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxn = 105;
const int maxm = 100005;
int v;
int dp[maxm];

void ZeroOnePack(int cost,int weight) {
for(int i = v; i >= cost; i--)
dp[i] = max(dp[i],dp[i - cost] + weight);
}

void CompletePack(int cost,int weight) {
for(int i = cost; i <= v; i++)
dp[i] = max(dp[i],dp[i - cost] + weight);
}

void MultiplePack(int cost,int weight,int n) {
if(cost * n > v) {
CompletePack(cost,weight);
} else {
int k = 1;
while(k < n) {
ZeroOnePack(cost * k,weight * k);
n -= k;
k *= 2;
}
ZeroOnePack(cost * n,weight * n);
}
}
int value[maxn];
int number[maxn];

bool Do() {
int n,m;
scanf("%d%d",&n,&m);
if(n == 0 && m == 0)
return false;

for(int i = 1;i <= n;i++)
scanf("%d",&value[i]);
for(int i = 1;i <= n;i++)
scanf("%d",&number[i]);

v = m;
memset(dp,0,sizeof(dp));

for(int i = 1;i <= n;i++) {
MultiplePack(value[i],value[i],number[i]);
}
int cnt = 0;
for(int i = 1;i <= m;i++) {
if(dp[i] != dp[i - 1])
cnt++;
}

printf("%d\n",cnt);
return true;
}

int main() {
while(Do());
return 0;
}

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