Hdu 3123 (0! + 1! + 2! + 3! + 4! + ... + n!)%m

Hdu 3123

Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m.

Constrains
0 < T <= 20
0 <= n < 10^100 (without leading zero)
0 < m < 1000000

提示： n！对于n大于m的时候 n!%m=0 因为n的阶乘包括m

#include<stdio.h>

#include<string.h>

int main()

{

int i,j,n,m,cas,t;

__int64 mid,sum;

char s[1000];

scanf("%d",&cas);

while(cas--)

{

scanf("%s %d",&s,&m);

if(strlen(s)>=7)

t=m;

else sscanf(s,"%d",&t);

mid=1;

sum=1;

for(i=1;i<=t;i++)

{

mid=(mid*i)%m;

sum=(sum+mid)%m;

}

printf("%I64d\n",sum%m);//这里一定一定要%m 否则会错

}

return 0;

}