hdu 1284 分硬币 && uva 147

#include<bits/stdc++.h>
using namespace std;
int main()
{
unsigned long long int dp[30005];
int money[]={10000,5000,2000,1000,500,200,100,50,20,10,5};
memset(dp,0,sizeof(dp));
dp[0]=1;
//cout<<money[10]<<endl;
for(int i=10;i>=0;i--)
{
for(int j=money[i];j<30005;j++)

{
dp[j]=dp[j-money[i]]+dp[j];
}
}
//cout<<dp[0]<<dp[5]<<dp[10]<<dp[15]<<endl;
double n;

//cin>>n;
//cout<<dp[5]<<endl;
//cout<<dp[10020]<<endl;
int a,b;
while((scanf("%d.%d",&a,&b))!=EOF&&a+b)
{
/*  n=n*100;
cout<<n;
int m=n;
printf("%6.2lf%17I64d\n",n/100,dp[m]);*/
int n=a*100+b;//cout<<n<<endl;
double ans=n*1.0/100.0;
printf("%6.2lf%17lld\n",ans,dp
);
}

return 0;
}

View Code

[b]uva 147[/b]

Description





New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1

20c, 2

10c, 10c+2

5c, and 4

5c.

Input

Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

Sample input

0.20
2.00
0.00

Sample output

0.20                4
2.00              293

dp[j]=dp[j]+dp[j-i],j表示钱数，i表示价值。首先i=1,这样得到不同的钱数被全部分为1的硬币的兑换法，然后把大于2的数分解为 2+i，即2+i的钱分为一个2加第i个钱全部
为1的分法，这么递推下去。