Big Number-Asia 2002, Dhaka (Bengal) （计算位数）题解

本文链接：https://blog.youkuaiyun.com/qq_14938523/article/details/79372008

本文介绍了一种计算大整数阶乘位数的方法，通过累加每个整数的log10值并取整加1得到结果。适用于密钥生成、加密等应用中需要处理非常大的整数的情况。

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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\n",(int)sum+1);
	}
	return 0;
}