HDU 3123 GCC (同余模定理)

 

GCC

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 3033    Accepted Submission(s): 950

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

 

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.

 

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

 

Sample Input

1

10 861017

 

Sample Output

593846

 

思路：n可以达到10^100，很明显不能直接处理。但是发现只要n>m，那么m!+(m+1)!+...+n!这些项都是可以被m整除的，要对m求余，只需要找比m小的阶乘即可，而m的范围为1000000，在O(m)的复杂度下是可以完成的。所以只需判断n是否大于m，若大于m，则取m-1即可，若小于m则不变。

 

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int T;
10     scanf("%d",&T);
11     while(T -- )
12     {
13         char N[1005];
14         int result = 0,m;
15         scanf("%s",N);
16         scanf("%d",&m);
17         if((int)strlen(N)>=7)
18             result = m - 1;
19         else
20         {
21             int mul = 1;
22             for(int i=(int)strlen(N)-1; i>=0; i--)
23             {
24                 result += ((N[i]-'0')*mul);
25                 mul *= 10;
26             }
27         }
28         if(result >= m)
29             result = m - 1;
30         __int64 temp = 1,ans = 1;
31         for(int i=1; i<=result; i++)
32         {
33             temp = (temp * (__int64)i) % m;
34             ans = (ans + temp) % m;
35         }
36         ans %= m;
37         printf("%I64d\n",ans);
38     }
39     return 0;
40 }