1424: Coin Change
| Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
|---|---|---|---|---|---|
| 3s | 8192K | 528 | 166 | 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 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
11 26
Sample Output
4 13
This problem is used for contest: 81 148
#include<stdio.h>
int main()
{
int coin[5]={1,5,10,25,50};
int a[7490]={0};
int i,j,n;
a[0]=1;
for(i=0;i<5;i++)
for(j=coin[i];j<7490;j++)
a[j]+=a[j-coin[i]];
while(scanf("%d",&n)!=-1)
printf("%d/n",a[n]);
return 0;
}
本文探讨了一种算法解决方案,用于计算特定金额下不同硬币组合的数量。通过使用五种面额的硬币(50美分、25美分、10美分、5美分和1美分),程序能够高效地找出任意金额(最高可达7489美分)的所有可能找零方式。
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