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hdu 3123
GCC
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3628 Accepted Submission(s): 1186
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
Source
2009 Asia Wuhan Regional Contest Online
1 /**开开心心找规律**/ 2 3 #include<iostream> 4 #include<stdio.h> 5 #include<cstdlib> 6 #include<cstring> 7 using namespace std; 8 typedef __int64 LL; 9 10 void solve(LL n,LL p) 11 { 12 LL i,j=1,s=1; 13 for(i=1;i<=n;i++) 14 { 15 j=(j*i)%p; 16 s=(s+j)%p; 17 } 18 printf("%I64d\n",s); 19 } 20 int main() 21 { 22 LL T; 23 char a[10000]; 24 LL i,n,m; 25 scanf("%I64d",&T); 26 while(T--) 27 { 28 scanf("%s%I64d",a,&n); 29 if(n==1){ 30 printf("0\n"); 31 continue; 32 } 33 for(i=0,m=0;a[i]!=‘\0‘;i++) 34 { 35 m=m*10+a[i]-‘0‘; 36 if(m>=n) 37 { 38 m=n; 39 break; 40 } 41 } 42 solve(m,n); 43 } 44 return 0; 45 }
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