GCC
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3993 Accepted Submission(s): 1304
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
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.
Each test on a single consists of two integer n and m.
Output
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
Constrains
0 < T <= 20
0 <= n < 10^100 (without leading zero)
0 < m < 1000000
Sample Input
1
10 861017
Sample Output
593846求 n!%m=(n-1)!*n%m; 即f(n)=f(n-1)*n%m;并且 (0!+1!+...n!)%m 假设n>m 那么以后的k!都能够被m整除。。忽略然后 。。递推就能够了#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define ll long long
using namespace std;
const int INF=1<<27;
const int maxn=1010;
int main()
{
int t;
char n[200];int m;
scanf("%d",&t);
while(t--)
{
scanf("%s %d",n,&m);
int tem;
if(strlen(n)>=7)
tem=999999;
else
sscanf(n,"%d",&tem);
tem=min(m,tem);
ll f1=1%m,fn,ans=f1;
for(int i=1;i<=tem;i++)
{
fn=f1*i%m;
ans+=fn;
f1=fn;
}
printf("%lld\n",ans%m);
}
return 0;
}