hdu5429（BestCoder Round #54 (div.2) 1003题）

Geometric Progression

Accepts: 40

Submissions: 644

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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 integerT(T≤20) which denotes the number of test cases. For each test case, there is an positive integern(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

41031 1 131 4 2516 8 4 2 1

Sample Output

YesYes

分析：大数，直接java过，判断条件a[i]*a[i]==a[i-1]*a[i+1]，另外要注意n个数全为0或者一部分为0的情况。

import java.util.*;import java.math.*;public class Main {    /**     * @param args     */    public static void main(String[] args) {        // TODO Auto-generated method stub        BigDecimal[] s = new BigDecimal[110];        int T,n,flag,i,t;        BigDecimal a,b;        Scanner cin = new Scanner(System.in);        T = cin.nextInt();        while (T-->0){            n = cin.nextInt();            flag = 0;            t = 0;            for (i=0; i<n; i++){                s[i] = cin.nextBigDecimal();                if (s[i].compareTo(BigDecimal.valueOf(0))==0){                    t++;                }            }            if (t==n||n==1){                System.out.println("Yes");                continue;            }            if (t!=0){                System.out.println("No");                continue;            }                            for (i=1; i<n-1; i++){                a = s[i].multiply(s[i]);                b = s[i-1].multiply(s[i+1]);                if (a.compareTo(b)!=0){                    flag = 1;                    break;                }            }            if (flag==1)                System.out.println("No");            else                System.out.println("Yes");        }    }}