这里给出我写的代码，这个代码无法通过所有的评测点，大部分运行超时
long long cnm(int n, int m)
{
int i;
long long sum = 1;
int k = 1;
if (m > n / 2)
m = n - m;
for ( i = n - m + 1; i <= n; i++)
{
sum *= (long long)i;
while (k <= m && sum % k == 0)
{
sum /= (long long)k;
k++;
}
}
return sum;
}
#include
int main()
{
int i, n, j, count = 0;
scanf("%d", &n);
for (i = 2; i <= 100000000; i++)
{
for (j = 0; j <= i; j++)
{
count++;
if (cnm(i, j) == n)
goto begin;
}
}
begin: if(n>1)printf("%d", count+3);
else
printf("1");
return 0;
}
如需正确答案，请自行搜索
