(a+b)%m=[(a%m)+(b%m)]%m
(a*b)%m=[(a%m)*(b%m)]%m
所以(0!+1!+2!+.....+n!)%m=[(0!%m+1!%m)%m+(1!%m)*(2%m)%m}%m.........到K>=n时，K!%m=0
所以分为n>=m和n<m两种情况
#include <stdio.h>
#include <string.h>
#include <math.h>
char s[101];
int main()
{
int t, n, m;
scanf("%d",&t);
while (t--)
{
scanf("%s %d", s, &m);
long long a = 1, b = 1;
int n = 0, i;
for (i=strlen(s)-1; i>=0; i--)
{
n += (s[i]-'0')*b;
if (n>=m)
break;
b *= 10;
}
b = 1;
if (n<m)
for (i=1;i<=n; i++)
{
b = (b*i)%m;
a = (a+b)%m;
}
else
for (i=1; i<m; i++)
{
b = (b*i)%m;
a = (a+b)%m;
}
printf("%I64d\n", a%m);
}
system("pause");
return 0;
}