Coins

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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. 

Sample Input
3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0
Sample Output
8

4

题意：给定不同面值的钱和数目，用这些钱组合成小于等于M元钱的钱数，记录能够组合成的钱数数目。

思路：完全背包。

依次遍历每种面值的钱，该面值钱的数目依次增加，在前计算得出的总钱数的基础上看能否组成新的总钱数，若能则该总钱数标记为1，否则保持初值0.

#include<cstdio>#include<cstring>using namespace std;int main(){    int n,m;    while(scanf("%d %d",&n,&m)&&(n||m))    {        int sum[m+1];        int cnt[m+1];//cnt记录当前状态的硬币个数        int a[n],c[n];        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<n;i++)            scanf("%d",&c[i]);        int cnts=0;//记录能够组合成的种类数        memset(sum,0,sizeof(sum));        sum[0]=1;        for(int i=0;i<n;i++)        {            memset(cnt,0,sizeof(cnt));            for(int j=a[i];j<=m;j++)            {                if(!sum[j]&&sum[j-a[i]]&&cnt[j-a[i]]<c[i])                {                    sum[j]=1;//能构成j元钱，则标记为1；                    cnt[j]=cnt[j-a[i]]+1;//硬币个数++；                    cnts++;//种类数++；                }            }        }        printf("%d\n",cnts);    }    return 0;}