【背包专题】 D

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.

InputThe 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.OutputFor each test case output the answer on a single line.Sample Input

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

Sample Output

8
4

题意：问能够将输入的钱币组合为1~m之间的数的种类总数，比如样例2：m=5，钱币可以组合为：1,2,4,5.所以总的种类数为4.
思路：在这道题中我们可以把一定数量组合而成的钱币总价值（1~m）看多背包费用和价值，最后判断费用和价值相等即记入总数。
还可以把每种钱币对应的每个数量计算出来，当作01背包来写这道题：http://blog.csdn.net/hello_sheep/article/details/77581462

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 100010
int dp[N],value[110],num[110];
int n,m,ans;
void CompletePack(int value,int cost)
{
    int i;
    for(i = cost; i <= m; i ++)
        dp[i] = max(dp[i],dp[i-cost]+value);
    return;
}
void ZeroOnePack(int value,int cost)
{
    int i;
    for(i = m; i >= cost; i --)
        dp[i] = max(dp[i],dp[i-cost]+value);
    return;
}
void MultiplePack(int num,int value,int cost)
{
    int k;
    if(value*num > m)
        CompletePack(value,cost);
    else
    {
        k = 1;
        while(k < num)
        {
            ZeroOnePack(k*value,k*cost);
            num -= k;
            k*=2;//为什么用左移运算符就tle!! 
        }
        if(num)
            ZeroOnePack(num*value,num*cost);
    }
    return;
}

int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),m+n)
    {
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= n; i ++)
            scanf("%d",&value[i]);
        for(i = 1; i <= n; i ++)
            scanf("%d",&num[i]);
            
        for(i = 1; i <= n; i ++)
            MultiplePack(num[i],value[i],value[i]);
        ans = 0;
        for(i = 1; i <= m; i ++)
            if(dp[i] == i)//费用和价值相等，满足条件 
                ans++;
        printf("%d
",ans);
    }
    return 0;
}