1424:Coin Change
| Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
|---|---|---|---|---|---|
| 3s | 8192K | 563 | 181 | Standard |
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 agiven amount of money.
For example, if we have 11 cents, then we can make changes with one10-cent coin and one 1-cent coin, two 5-cent coins and one 1-centcoin, 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 theabove coins. Note that we count that there is one way of makingchange for zero cent.
Write a program to find the total number of different ways ofmaking changes for any amount of money in cents. Your programshould be able to handle up to 7489 cents.
Input
The input file contains any number of lines, each one consisting ofa number for the amount of money in cents.Output
For each input line, output a line containing the number ofdifferent ways of making changes with the above 5 types of coins.Sample Input
11
26
Sample Output
#include<stdio.h>
#include<string.h>
int c[5]={50,25,10,5,1};
int v;
unsigned long f[8000];
int main()
{
memset(f,0,sizeof(f));
f[0]=1;
for(int i=0;i<5;i++)
for(int j=c[i];j<=7500;j++)
{
f[j]+=f[j-c[i]];//重要
}
while(scanf("%d",&v)==1)
{
printf("%dn",f[v]);
}
return 0;
}
本文介绍了一种解决硬币找零问题的算法实现,通过动态规划的方法找出特定金额下所有可能的找零组合数。输入为任意金额,输出为使用五种面额硬币的所有不同找零方式的数量。
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