POJ-1742：Coins（多重背包二进制求法）

Coins

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 38612 Accepted: 13100

Description

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

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

84

思路：多重背包（二进制形式）。注意：不要使用max函数！！！！会TLE！！

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;int v[105],c[105];int d[100001];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0&&m==0)break;        for(int i=0;i<n;i++)scanf("%d",&v[i]);        for(int i=0;i<n;i++)scanf("%d",&c[i]);        memset(d,0,sizeof d);        for(int i=0;i<n;i++)        {            if(c[i]*v[i]>=m)            {                for(int j=v[i];j<=m;j++)                {                    if(d[j-v[i]]+v[i]>d[j])d[j]=d[j-v[i]]+v[i];                }            }            else            {                int k=1;                while(c[i]-k>0)                {                    for(int j=m;j>=v[i]*k;j--)                    {                        if(d[j-v[i]*k]+v[i]*k>d[j])d[j]=d[j-v[i]*k]+v[i]*k;                    }                    c[i]-=k;                    k*=2;                }                for(int j=m;j>=v[i]*c[i];j--)                {                    if(d[j-v[i]*c[i]]+v[i]*c[i]>d[j])d[j]=d[j-v[i]*c[i]]+v[i]*c[i];                }            }        }        int ans=0;        for(int i=1;i<=m;i++)ans+=(d[i]==i);        printf("%d\n",ans);    }    return 0;}