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  • HDOJ 2069 Coin Change(母函数)

    Coin Change

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 10289    Accepted Submission(s): 3451


    Problem Description
    Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.

    For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.

    Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.
     
    Input
    The input file contains any number of lines, each one consisting of a number ( ≤250 ) for the amount of money in cents.
     
    Output
    For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.
     
    Sample Input
    11
    26
     
    Sample Output
    4
    13
     
    Author
    Lily
     
    Source
     
    Recommend
    linle
     
     1 //加一维限制数量。。。。。。
     2 
     3 #include <iostream>
     4 #include <cstdio>
     5 #include <cstring>
     6 
     7 using namespace std;
     8 
     9 const int value[6]={0,1,5,10,25,50};
    10 
    11 int c1[300][110],c2[300][110];
    12 int n;
    13 
    14 int main()
    15 {
    16 while(scanf("%d",&n)!=EOF)
    17 {
    18    memset(c1,0,sizeof(c1));
    19    memset(c2,0,sizeof(c2));
    20 
    21    for(int i=0;i<=min(n,100);i++)
    22    {
    23         c1[i][i]=1;
    24    }
    25 
    26    for(int i=2;i<=5;i++)
    27    {
    28        for(int j=0;j<=n;j++)
    29        {
    30            for(int k=0;k+j<=n;k+=value[i])
    31            {
    32                for(int l=0;l<=100&&l+k/value[i]<=100;l++)
    33                {
    34                    c2[j+k][l+k/value[i]]+=c1[j][l];
    35                }
    36            }
    37        }
    38 
    39        for(int j=0;j<=n;j++)
    40        {
    41            for(int k=0;k<=100;k++)
    42            {
    43                c1[j][k]=c2[j][k];
    44                c2[j][k]=0;
    45            }
    46        }
    47    }
    48 
    49    int ans=0;
    50    for(int j=0;j<=100;j++)
    51    {
    52        ans+=c1[n][j];
    53    }
    54 
    55    printf("%d
    ",ans);
    56 
    57 }
    58     return 0;
    59 }
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  • 原文地址:https://www.cnblogs.com/CKboss/p/3165222.html
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