GCC
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3754 Accepted Submission(s):
1216
Problem Description
The GNU Compiler Collection (usually shortened to GCC)
is a compiler system produced by the GNU Project supporting various programming
languages. But it doesn’t contains the math operator “!”.
In mathematics the
symbol represents the factorial operation. The expression n! means "the product
of the integers from 1 to n". For example, 4! (read four factorial) is 4 × 3 × 2
× 1 = 24. (0! is defined as 1, which is a neutral element in multiplication, not
multiplied by anything.)
We want you to help us with this formation: (0! + 1!
+ 2! + 3! + 4! + ... + n!)%m
is a compiler system produced by the GNU Project supporting various programming
languages. But it doesn’t contains the math operator “!”.
In mathematics the
symbol represents the factorial operation. The expression n! means "the product
of the integers from 1 to n". For example, 4! (read four factorial) is 4 × 3 × 2
× 1 = 24. (0! is defined as 1, which is a neutral element in multiplication, not
multiplied by anything.)
We want you to help us with this formation: (0! + 1!
+ 2! + 3! + 4! + ... + n!)%m
Input
The first line consists of an integer T, indicating the
number of test cases.
Each test on a single consists of two integer n and
m.
number of test cases.
Each test on a single consists of two integer n and
m.
Output
Output the answer of (0! + 1! + 2! + 3! + 4! + ... +
n!)%m.
n!)%m.
Constrains
0 < T <= 20
0 <= n < 10^100 (without
leading zero)
0 < m < 1000000
Sample Input
1
10 861017
Sample Output
593846
Source
数字看起来非常巨大，吓人。
但是如果n!%p =0 那么 (n+1)!%p =0;
就这样。
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef __int64 LL; void solve(int sum,int p)
{
int i;
LL ans = ,cur = % p ;
for(i=;i<=sum;i++)
{
ans = (ans * i)%p;
cur = (cur + ans)%p;
if(ans == )break;
}
printf("%I64d\n",cur);
}
int main()
{
int T;
int i,p;
char a[];
scanf("%d",&T);
while(T--)
{
scanf("%s%d",a,&p);
if(p==)
{
printf("0\n");
continue;
}
int sum = ;
for(i=;a[i]!='\0';i++)
{
sum=sum*+a[i]-'';
if(sum>=p)break;
}
solve(sum,p);
}
return ;
}