hdu 2069 Coin Change 背包。本来打算用母函数再写一遍的，发现代码极其相似，就没写

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14982    Accepted Submission(s): 5070

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

1126

 

Sample Output

413

 

Author

Lily

 

Source

浙江工业大学网络选拔赛

 

由于最多只能用100枚硬币，所以数组得再加一维

dp[i][j]代表的是总额为i需要j枚硬币。

dp[i][j] = dp[i-v][j-1] ；

代码；

#include <stdio.h>#include <string.h>#define MAX 300#define COUNT 100const int type[]={50,25,10,5,1} ;int dp[MAX][110] ;int max(int a , int b){return a>b?a:b ;}int main() {dp[0][0] = 1 ;for(int i = 0 ; i < 5 ; ++i){for(int j = type[i] ; j <= 250 ; ++j){for(int m = 0 ; m < 100 ; ++m){dp[j][m+1] += dp[j-type[i]][m] ;}}}int n ;while(~scanf("%d",&n)){int ans = 0 ;for(int i = 0 ; i <= 100 ; ++i)ans += dp[n][i] ;printf("%d\n",ans) ;}return 0 ;}

与君共勉