Eddy's AC難題
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2584 Accepted Submission(s): 1211
特別說明:爲了問題的簡化,我們這裏假設摘錄下的人數爲n人,而且每個人ac的數量不會相等,最後結果在64位整數範圍內.
#include <stdio.h>
__int64 group(int m,int n)//////判斷有多少組合
{
int i;
__int64 s=m-1;
if(n-m<m)
m=n-m;
for(i=1;i<=m;i++)
{
s*=(n-i+1);
s/=i;
}
return s;
}
int main()
{
int i,n;
__int64 s;
while(scanf("%d",&n)!=EOF)
{
s=0;
for(i=2;i<=n;i++)
s+=group(i,n);
printf("%I64d\n",s);
}
return 0;
}