hdu 2069Coin Change

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博客围绕硬币找零问题展开，给定5种硬币（50分、25分、10分、5分、1分），要求编写程序计算为任意金额找零的不同方式总数，程序需能处理最多100枚硬币，还给出了输入输出格式及示例。

Coin Change

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25654 Accepted Submission(s): 8971

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

Sample Output

#include <bits/stdc++.h>

using namespace std;
const int COIN = 101;
const int MONEY = 251;
int dp[MONEY][COIN]={0};
int type[5] = {1, 5, 10, 25, 50};
int ans[MONEY];
void slove()
{
    dp[0][0] = 1;
    for(int i = 0; i < 5; i++)
    {
        for(int j = 1; j < COIN; j++)
        {
            for(int k = type[i]; k < MONEY; k++)
            {
                dp[k][j] += dp[k-type[i]][j-1];
            }
        }
    }

    for(int i = 0; i < MONEY; i++)
    {
        for(int j = 0; j < COIN; j++)
            ans[i] += dp[i][j];
    }
}
int main()
{
  int n;
    slove();
  while(~scanf("%d", &n))
  {
    printf("%d\n",ans[n]);
  }
}