BestCoder Round #54 (div.2) HDU5329 Geometric Progression
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Geometric Progression
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1040 Accepted Submission(s): 294
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 powersrk 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.
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
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 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.
For each test case, there is an positive integer
Output
For each case, output "Yes" or "No".
Sample Input
41031 1 131 4 2516 8 4 2 1
Sample Output
YesYesNoYes
Source
BestCoder Round #54 (div.2)
出题人:判断是否为等比数列,可以检验对所有1<i<n
A[i−1]∗A[i+1]=A[i]∗A[i] 是否都成立。
直接高精度也是资词的。比较简单的方法是选择若干质数
(保证乘积大于10200),在模意义下检验。
复杂度O(k∗n)。k表示选取的质数个数。
1003:HDU5329 Geometric Progression
Geometric Progression
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1040 Accepted Submission(s): 294
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 powersrk 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.
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
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 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.
For each test case, there is an positive integer
Output
For each case, output "Yes" or "No".
Sample Input
41031 1 131 4 2516 8 4 2 1
Sample Output
YesYesNoYes
Source
BestCoder Round #54 (div.2)
出题人:判断是否为等比数列,可以检验对所有1<i<n
A[i−1]∗A[i+1]=A[i]∗A[i] 是否都成立。直接高精度也是资词的。
比较简单的方法是选择若干质数(保证乘积大于10200),
在模意义下检验。复杂度O(k∗n)。k表示选取的质数个数。
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