题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2069
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
解题思路：这道题可以用暴力枚举直接解决。枚举每种硬币的数量为0~n/这种币值即可，为了避免TLE超时，最后一种硬币换成表达式来判断，用num来计数情况，其中注意所有币值的总数量<=100。
AC代码一(直接暴力)：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,num;
while(cin>>n){
num=;
for(int a=;a*<=n;a++){
for(int b=;b*<=n;b++){
for(int c=;c*<=n;c++){
for(int d=;d*<=n;d++){//剩下一步由减法来,避免超时
if(n-a*-b*-c*-d*>= && a+b+c+d+n-a*-b*-c*-d*<=)num++;
}
}
}
}
cout<<num<<endl;
}
return ;
}
AC代码二之dp：先贴一下此题的思路：题解报告：hdu 1284 钱币兑换问题（简单数学orDP）这题就是多加了一个维度，因为题目中规定了硬币的数量最多取100个，因此定义dp[k][j]表示前k个硬币组成钱j的总方案数，那么易得状态转移方程：dp[k][j]+=dp[k-1][j-a[i]]，意思是减去当前某种一个币值，那么就会增加前k-1个硬币组成钱j-a[i]的方案数dp[k-1][j-a[i]](对于同一种硬币a[i]来讲)。预处理打表，然后累加用k（k∈[0,100]）个硬币组成钱n的所有方案数即为最终的方案总数。注意：①初始化dp[0][0]=1，表示前0个硬币组成钱0的方案数为1（原因和上面链接博文里的一样）答案要累加不超过i:0-->100得到的总方案数,一定要从0开始累加，因为有0个硬币时的方案数。
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,sum,a[]={,,,,},dp[][];
memset(dp,,sizeof(dp));dp[][]=;
for(int i=;i<;++i)//种数
for(int k=;k<=;++k)//硬币总数不超过100
for(int j=a[i];j<=;++j)
dp[k][j]+=dp[k-][j-a[i]];
while(cin>>n){
sum=;
for(int k=;k<=;++k)sum+=dp[k][n];//累加用0~100组成钱n的所有方案数
cout<<sum<<endl;
}
return ;
}