zoukankan      html  css  js  c++  java
  • Big Number-Asia 2002, Dhaka (Bengal) (计算位数)题解

    Big Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 40782    Accepted Submission(s): 19958


    Problem Description
    In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.
     

    Input
    Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 107 on each line.
     

    Output
    The output contains the number of digits in the factorial of the integers appearing in the input.
     

    Sample Input
    2 10 20
     

    Sample Output
    7 19


    思路:

    原来还有这种操作:求n的位数为(int)log10(n)+1,阶乘为相加后加1

    code

    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<cctype>
    #include<queue>
    #include<math.h>
    #include<iostream>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    #define N 85
    using namespace std;
    int main(){
    	int t,n,i;
    	double sum;
    	scanf("%d",&t);
    	while(t--){
    		scanf("%d",&n);
    		sum=0;
    		for(i=1;i<=n;i++){
    			sum+=log10(i);
    		}
    		printf("%d
    ",(int)sum+1);
    	}
    	return 0;
    }


  • 相关阅读:
    mysql8 JDBC连接注意事项
    Flask上下文管理源码分析
    python基础学习day5
    python基础学习day4
    python基础学习day3
    python基础学习day2
    python基础学习day1
    简单的名片管理系统
    pycharm教程
    Python:知识目录
  • 原文地址:https://www.cnblogs.com/KirinSB/p/9409134.html
Copyright © 2011-2022 走看看