1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=615
2、题目大意：
现在有五种硬币，面值分别是1，5,10,25,50，现在给定 一个金额值，将其换成以上的硬币，求有多少种换法
3‘、想了好久终究是没有想对状态转移方程
正确的方法应该是：dp[i][j]表示用前j种硬币组成i分钱时的种类数，那么状态转移方程是
dp[i][j]+=DP(i-k*v[j],j-1)
4、题目：
Coin Change
| Coin Change |
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
5、ac代码：
//dp[i][j]表示用前j种硬币组成i分的种类数
#include<stdio.h>
#include<string.h>
#define N 7500
int dp[N][6];
int v[5]={1,5,10,25,50};
int DP(int i,int j)
{if(j==0)return dp[i][j]=1;if(dp[i][j]!=-1)return dp[i][j];dp[i][j]=0;for(int k=0;i-k*v[j]>=0;k++){dp[i][j]+=DP(i-k*v[j],j-1);}return dp[i][j];
}
int main()
{int n;memset(dp,-1,sizeof(dp));//memset只需初始一次即可，如果后边n很大，可以用之前存好的dp[][]while(scanf("%d",&n)!=EOF){ //memset(dp,-1,sizeof(dp));//如果每次都初始化，那么每次都会从头算，会超时for(int i=0;i<=n;i++)dp[i][0]=1;printf("%d\n",DP(n,4));}return 0;
}