POJ1742--Coins(动态规划)

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

#include<iostream>
#include<algorithm>
#include<string.h>
int dp[][100003];
int val[],num[];
using namespace std;
int main(){
    int n,m;
    while(cin>>n>>m&&!(n==&&m==)){
        memset(dp,,sizeof(dp));
        for(int i=;i<n;i++){
            cin>>val[i];
        }
        for(int i=;i<n;i++){
            cin>>num[i];
        }
        dp[][]=;
        for(int i=;i<n;i++){
            for(int j=;j<=m;j++){
                for(int w=;w<=num[i]&&w*val[i]<=j;w++){
                    dp[i+][j]|=dp[i][j-w*val[i]];
                }
            }
        }
        int ans=count(dp[n]+,dp[n]++m,);
        cout<<ans<<endl;
    }
    return ;
} 

dp[i+1][j]表示前i种数字能否拼成j
一般用DP求取bool结果的话会有不少浪费，同样的复杂度可以获得很多信息

优化

dp[i+1][j]:用前i种数加和得到j时第i种数最多能剩几个

1. dp[i][j] := 用前i种硬币凑成j时第i种硬币最多能剩余多少个（-1表示配不出来）
2. 如果dp[i - 1][j] >= 0(前i-1个数可以凑出j，那么第i个数根本用不着)直接为C[i]
3. dp[i][j] =  如果j < A[i]或者dp[i][j - a[i]] <=0 (面额太大或者在配更小的数的时候就用光了）-1
4. 其他（将第i个数用掉一个） dp[i][j-a[i]] - 1

#include<iostream>
#include<algorithm>
#include<string.h>
int dp[][100003];
int val[],num[];
using namespace std;
int main(){
    int n,m;
    while(cin>>n>>m&&!(n==&&m==)){
        memset(dp,-,sizeof(dp));
        for(int i=;i<n;i++){
            cin>>val[i];
        }
        for(int i=;i<n;i++){
            cin>>num[i];
        }
        dp[][]=;
        for(int i=;i<n;i++){
            for(int j=;j<=m;j++){
                if(dp[i][j]>=)
                    dp[i+][j]=num[i];
                else if(j<val[i]||dp[i+][j-val[i]]<=){
                    dp[i+][j]=-;
                }
                else{
                    dp[i+][j]=dp[i+][j-val[i]]-;
                }
            }
        }
        int ans=;
        for(int i=;i<=m;i++){
            if(dp[n][i]!=-)
                ans++;
        }
        cout<<ans<<endl;
    }
    return ;
} 

数组重复利用

#include<iostream>
#include<algorithm>
#include<string.h>
int dp[];
int val[],num[];
using namespace std;
int main(){
    int n,m;
    while(cin>>n>>m&&!(n==&&m==)){
        memset(dp,-,sizeof(dp));
        for(int i=;i<n;i++){
            cin>>val[i];
        }
        for(int i=;i<n;i++){
            cin>>num[i];
        }
        dp[]=;
        for(int i=;i<n;i++){
            for(int j=;j<=m;j++){
                if(dp[j]>=)
                    dp[j]=num[i];
                else if(j<val[i]||dp[j-val[i]]<=){
                    dp[j]=-;
                }
                else{
                    dp[j]=dp[j-val[i]]-;
                }
            }
        }
        int ans=;
        for(int i=;i<=m;i++){
            if(dp[i]!=-)
                ans++;
        }
        cout<<ans<<endl;
    }
    return ;
}