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  • hdoj--3123--GCC(技巧阶乘取余)

    GCC

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 4473    Accepted Submission(s): 1472



    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
    2009 Asia Wuhan Regional Contest Online

    好吧,0!是1,我服了,,,,,

    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<algorithm>
    #define ll long long
    #define N 1000010
    using namespace std;
    char s[110];
    ll sum[N];
    ll change(char *s)
    {
    	int l,i,ans=0;
    	l=strlen(s);
    	for(i=0;i<l;i++)
    		ans=ans*10+(s[i]-'0');
    	return ans; 
    }
    int main()
    {
    	int t;
    	int n,m;
    	int i,j,k,l;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%s%d",s,&m);
    		l=strlen(s);
    		if(l<7)
    		{
    			n=change(s);
    			if(n>m)
    				n=m-1;
    			sum[0]=1;
    			int ans=0;
    			for(i=1;i<=n;i++)
    			{
    				sum[i]=(sum[i-1]*i)%m;
    				ans=(ans+sum[i])%m;
    			}
    			printf("%d
    ",(ans+1)%m);
    		}
    		else
    		{
    			n=m-1;
    			sum[0]=1;
    			int ans=0;
    			for(i=1;i<=n;i++)
    			{
    				sum[i]=(sum[i-1]*i)%m;
    				ans=(ans+sum[i])%m;
    			}
    			printf("%d
    ",(ans+1)%m);
    		}
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273629.html
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