hdoj--2069--Coin Change(动态规划)
2024-08-31 09:41:54
Coin Change
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16032 Accepted Submission(s): 5441
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.
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[500][600];
int num[1000];
int coin[6]={0,1,5,10,25,50};
void getnum()
{
memset(dp,0,sizeof(dp));
memset(num,0,sizeof(num));
dp[0][0]=1;
for(int i=1;i<=5;i++)
{
for(int j=0;j<=250;j++)//货币总面额不超过250
{
for(int k=0;k<100;k++)//硬币总数不超过100个
{
if(coin[i]+j<=250)
dp[coin[i]+j][k+1]+=dp[j][k];//当前货币总面值为j,分成k个硬币
else
break;
}
}
}
for(int i=0;i<=250;i++)
{
for(int j=0;j<=100;j++)
num[i]+=dp[i][j];
}
}
int main()
{
int n;
getnum();
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",num[n]);
}
}
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