674 - Coin Change

来源:互联网 发布:单机版进销存软件 编辑:程序博客网 时间:2024/06/05 14:51

  Coin Change 

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, two 5-cent coins and one 1-cent coin, 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 7489 cents.

Input 

The input file contains any number of lines, each one consisting of a number 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 

1126

Sample Output 

413



Miguel Revilla 
2000-08-14

/*动态规划,具体看代码就会明白啦,不过要注意的是,dp一次就要好*/#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int N=7490;int dp[N];int main(){int coins[5]={1,5,10,25,50};int n;int i,j;dp[0]=1;for(i=0;i<5;i++){for(j=0;j<=N-coins[i];j++){if(dp[j]!=0){dp[j+coins[i]]+=dp[j];}}}while(scanf("%d",&n)!=EOF){cout<<dp[n]<<endl;}return 0;}


0 0
原创粉丝点击