现在的翻译其本上都是这样的，我也没有办法啊！呵呵！

也就是说 假如 是F的时候 那么 是否小于250的时候又该另外考虑呢？ 最近没上这个论坛 不好意思

#include <stdio.h>
#define N 20
void main()
{
int n[N],k[N],i,j,count=0;
double sum1=1.0,sum2=1.0;
for(i=0;i<N;i++)
{
scanf("%d,%d",&n[i],&k[i]);
if(n[i]==0&&k[i]==0)
break;
count++;
}
for(j=0;j<count;j++)
{
sum1=1;
sum2=1;
for(i=1;i<=k[j];i++)
{
sum1=sum1*n[j];
n[j]--;
}
for(i=1;i<=k[j];i++)
{
sum2=sum2*i;
}
printf("%-10.0f\n",sum1/sum2);
}
}
我做的第二题

这个好像是不对吧！楼上的‘小小贝’当你输入 49 6 时输出的结果是什么呀？呵呵，好像和问题不一样啊！

没错啊.....就是你要得结果啊....
输入时要49,6这样输入的.你不会输错吧

晕。想念起了搞ACM的时光

printf("%-10.0f\n",sum1/sum2);
为什么这样打印出的是整数啊？

sum1*1.0 / sum2*1.0

谢谢，我知道了！原来是这么一回事啊！

The Angle
The interval angle between the hour hand and the minute hand is always among 0 degree and 180 degree(including 0 and 180 degree). For example, when it's 12 o'clock, the angle of the two hands is 0 while 6:00 is 180 degree. Try to calculate any degree when it's between 12:00 to 11:59.

Input
The input data are of various situations. Every team consists of two numbers : the first number stands for the hour(great than 0 and less than or equal to 12) and second represents the minute( among [0, 59]). The input ends when the two numbers are both zeros.

Output
Print out the minimum angle between the two hands with the normal time form outputed.

Input Sample
12 012 306 03 00 0 Output Sample
At 12:00 the angle is 0.0 degrees.
At 12:30 the angle is 165.0 degrees.
At 6:00 the angle is 180.0 degrees.
At 3:00 the angle is 90.0 degrees.