GCC

最新推荐文章于 2020-11-30 16:26:54 发布

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最新推荐文章于 2020-11-30 16:26:54 发布

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文章标签：操作系统

原文链接：http://www.cnblogs.com/LK1994/p/3405467.html

本文介绍了一个基于GCC编译器的数学运算挑战问题，探讨了如何计算从0到n的阶乘之和并对m取余的算法实现。通过分析题目特点，提出了一种高效的分步取余方法。

题目描述

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.

输入

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.

0 < T <= 20
0 <= n < 10¹⁰⁰(without leading zero)
0 < m < 1000000

输出

Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m.

示例输入

1 
10 861017

示例输出

593846

题意：输入n和m，计算(0! + 1! + 2! + 3! + 4! + ... + n!)%m的值。。。
思路：因为看到 n 最大是10^100，顿时感觉很蒙，可是，只要仔细想想，n这么大是唬人的。因为只要n>=m时，它的阶乘对m取余都是0，也就相当于 (0! + 1! + 2! + 3! + 4! + ... + （m-1)!)%m.
　　　若原本n就小于m，直接算就行了。就是分步取余。

 1 #include<stdio.h>
 2 #include<string.h>
 3 
 4 int MOD(int n,int m)
 5 {
 6     long long ans = 1,tmp = 1;
 7     int num;
 8     for(num = 1; num <= n; num++)
 9     {
10         tmp = (tmp*(num%m))%m;
11         ans = (ans+tmp)%m;
12     }
13     ans = ans%m;
14     return ans;
15 }
16 
17 int main()
18 {
19     int test;
20     char s[110];
21     int n,m,ans;
22     scanf("%d",&test);
23     while(test--)
24     {
25         scanf("%s %d",s,&m);
26         if(m == 1)
27         {
28             printf("0\n");
29             continue;
30         }
31         n = 0;
32         for(int i = 0; s[i]; i++)
33         {
34             n = n*10+s[i]-'0';
35             if(n >= m)
36             {
37                 n = m-1;
38                 break;
39             }
40         }
41         ans = MOD(n,m);
42         printf("%d\n",ans);
43     }
44     return 0;
45 }