UVA 147 Dollars

这道题要注意高精度误差，我这样做稍有一些麻烦，不如将面值放大成1, 2, 4, 10, 20, 40, 100, 200, 400, 1000, 2000再用整数计算，省去了我这个浮点误差（代码中的+0.1）的麻烦。

代码如下：

#include<stdio.h>
#include<string.h>
double n;
double chang[12] = { 0,0.05,0.1,0.2,0.5,1,2,5,10,20,50,100};
long long f[6000+10];
void solve()
{
    double amount = 300.10;
    memset(f,0,sizeof(f));
    f[0] = 1;
    for(int i = 1; i <= 11; i ++)
        for(double j = chang[i]; j <= amount; j += 0.05)
            f[(long long)(j/0.05+0.1)] += f[(long long)((j - chang[i])/0.05 +0.1)];
}
void input()
{
    solve();
    while(scanf("%lf",&n) == 1)
    {
        if(n == 0) break;
        printf("%6.2f%17lld\n",n,f[(long long)(n/0.05 +0.1)]);
    }
}
int main()
{
    input();
    return 0;
}