【Java大数】hdu 5429 Geometric Progression

Source : hdu 5429 Geometric Progression

http://acm.hdu.edu.cn/showproblem.php?pid=5429

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”.

题意

给n个数，判断能否构成等比数列（注意数的范围）

示例

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

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思路

大数据，我用的JAVA！

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参考代码

import java.io.*;import java.util.*;import java.math.*;public class Main{    static BigInteger []a = new BigInteger [110];    static BigInteger zero = BigInteger.ZERO;    public static void main(String[] args){        Scanner cin = new Scanner(System.in);        int T = cin.nextInt();        while(T-- >0){            int n = cin.nextInt();            boolean has0 = false,all0 = true;            for(int i = 0; i < n; ++ i){                a[i] = cin.nextBigInteger();                if(a[i].compareTo(zero) == 0) has0 = true;                else all0 = false;            }            if(all0 == true){System.out.println("Yes");continue;}            if(has0 == true){System.out.println("No");continue;}            boolean ok = true;            for(int i = 2; i < n; ++ i){                if(a[i-1].multiply(a[i-1]).equals(a[i-2].multiply(a[i])) == false)                {                    ok = false;                    break;                }            }            if(ok) System.out.println("Yes");            else System.out.println("No");        }    }}

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