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  • 动态规划(背包问题):POJ 1742 Coins

    Coins
    Time Limit: 3000MS   Memory Limit: 30000K
    Total Submissions: 32955   Accepted: 11199

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

    Sample Output

    8
    4

      
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 using namespace std;
     6 int num[110],v[110];
     7 int dp[2][100010];
     8 int main()
     9 {
    10     int n,m,now,pre;
    11     while(~scanf("%d%d",&n,&m)&&n&&m)
    12     {
    13         for(int i=1;i<=n;i++)scanf("%d",&v[i]);
    14         for(int i=1;i<=n;i++)scanf("%d",&num[i]);
    15         memset(dp,127,sizeof(dp));
    16         for(int i=1;i<=n;i++){
    17             now=i%2;pre=(i-1)%2;
    18             dp[now][0]=0;
    19             for(int j=1;j<=m;j++){
    20                 if(dp[pre][j]<=num[i-1])dp[now][j]=0;
    21                 if(j>=v[i]&&dp[now][j-v[i]]+1<=num[i])
    22                     dp[now][j]=min(dp[now][j],dp[now][j-v[i]]+1);    
    23             }
    24         }    
    25         
    26         int ans=0;
    27         for(int i=1;i<=m;i++)
    28             if(dp[now][i]<=num[n])
    29                 ans++;
    30         
    31         printf("%d
    ",ans);                    
    32     } 
    33     return 0;
    34 }
    尽最大的努力,做最好的自己!
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  • 原文地址:https://www.cnblogs.com/TenderRun/p/5262335.html
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