BestCoder Round #54 (div.2) HDOJ 5429 Geometric Progression(模拟)

Geometric Progression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 316    Accepted Submission(s): 86

Problem Description

Determine whether a sequence is a Geometric progression or not.

In mathematics, a **geometric progression**, also known as a **geometric sequence**, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.

Examples of a geometric sequence are powers rk of a fixed number r, such as 2k and 3k. The general form of a geometric sequence is

a, ar, ar2, ar3, ar4, …

where r ≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.

 

Input

First line contains a single integer T(T≤20) which denotes the number of test cases. 

For each test case, there is an positive integer n(1≤n≤100) which denotes the length of sequence,and next line has n nonnegative numbers Ai which allow leading zero.The digit's length of Ai no larger than 100.

 

Output

For each case, output "Yes" or "No".

 

Sample Input

4
1
0
3
1 1 1
3
1 4 2
5
16 8 4 2 1

 

Sample Output

Yes
Yes
No
Yes

 

等比数列a[i - 1] * a[i + 1] == a[i] * a[i]的应用。

由于是大数操作，所以选择Java语言，运用相关函数即可。

AC代码：

import java.io.*;
import java.util.*;
import java.math.BigInteger;
class Main {
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int t = cin.nextInt();
		for(int cas = 0; cas < t; ++cas) {
			int n = cin.nextInt();
			BigInteger a = cin.nextBigInteger();
			boolean allzero = true, nozero = true, ok = true;
			if(a.equals(BigInteger.ZERO) == false) allzero = false;
			else nozero = false;
			if(n == 1) {
				System.out.println("Yes");
				continue;
			}
			BigInteger b = cin.nextBigInteger();
			if(b.equals(BigInteger.ZERO) == false) allzero = false;
			else nozero = false;
			for(int i = 2; i < n; ++i) {
				BigInteger x = cin.nextBigInteger();
				if(x.equals(BigInteger.ZERO) == false) allzero = false;
				else nozero = false;
				if(ok)
					if(a.multiply(x).equals(b.multiply(b)) == false) ok = false;
				a = b;
				b = x;
			}
			if(allzero) System.out.println("Yes");
			else if(!nozero) System.out.println("No");
			else if(!ok) System.out.println("No");
			else System.out.println("Yes");
		}
	}
}