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  • hdu 3123

    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
     
    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
     
    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。
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<cstring>
     4 #include<cstdlib>
     5 using namespace std;
     6 typedef __int64 LL;
     7 
     8 void solve(int sum,int p)
     9 {
    10     int i;
    11     LL ans = 1 ,cur = 1 % p ;
    12     for(i=1;i<=sum;i++)
    13     {
    14         ans = (ans * i)%p;
    15         cur = (cur + ans)%p;
    16         if(ans == 0)break;
    17     }
    18     printf("%I64d
    ",cur);
    19 }
    20 int main()
    21 {
    22     int T;
    23     int i,p;
    24     char a[1002];
    25     scanf("%d",&T);
    26     while(T--)
    27     {
    28         scanf("%s%d",a,&p);
    29         if(p==1)
    30         {
    31             printf("0
    ");
    32             continue;
    33         }
    34         int sum = 0;
    35         for(i=0;a[i]!='';i++)
    36         {
    37             sum=sum*10+a[i]-'0';
    38             if(sum>=p)break;
    39         }
    40         solve(sum,p);
    41     }
    42     return 0;
    43 }
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  • 原文地址:https://www.cnblogs.com/tom987690183/p/3862550.html
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