题目来源

一、题目

A Pythagorean triple is a triple of integer numbers (a,b,c) such that it is possible to form a right triangle with the lengths of the first cathetus, the second cathetus and the hypotenuse equal to a, b and c, respectively. An example of the Pythagorean triple is (3,4,5).
Vasya studies the properties of right triangles, and he uses a formula that determines if some triple of integers is Pythagorean. Unfortunately, he has forgotten the exact formula; he remembers only that the formula was some equation with squares. So, he came up with the following formula: c=a^2−b.
Obviously, this is not the right formula to check if a triple of numbers is Pythagorean. But, to Vasya’s surprise, it actually worked on the triple (3,4,5): 5=3^2−4, so, according to Vasya’s formula, it is a Pythagorean triple.
When Vasya found the right formula (and understood that his formula is wrong), he wondered: how many are there triples of integers (a,b,c) with 1≤a≤b≤c≤n such that they are Pythagorean both according to his formula and the real definition? He asked you to count these triples.

给定n，问多少1≤a≤b≤c≤n，使直角三角形 (a,b,c) 满足c=a^2−b

二、输入

The first line contains one integer t (1≤t≤104) — the number of test cases.
Each test case consists of one line containing one integer n (1≤n≤109).

第一行测试样例数t
第二行整数n

三、输出

For each test case, print one integer — the number of triples of integers (a,b,c) with 1≤a≤b≤c≤n such that they are Pythagorean according both to the real definition and to the formula Vasya came up with.

满足条件的三角形数量

四、样例

input:

output:

0
1
1

Note
The only Pythagorean triple satisfying c=a2−b with 1≤a≤b≤c≤9 is (3,4,5); that’s why the answer for n=3 is 0, and the answer for n=6 (and for n=9) is 1.

五、思路 + 代码

联立 c2=a2+b2c^2=a^2+b^2c2=a2+b2 与 c=a2−bc=a^2-bc=a2−b 得 c2=b+c+b2c^2=b+c+b^2c2=b+c+b2
整理得 (c+b)(c−b)=b+c(c+b)(c-b)=b+c(c+b)(c−b)=b+c 即 c=b+1c=b+1c=b+1
将c=b+1c=b+1c=b+1代入c=a2−bc=a^2-bc=a2−b 得 b=a2−12b=\frac{a^2-1}{2}b=2a2−1

整理得:

a=aa=aa=a
b=a2−12b=\frac{a^2-1}{2}b=2a2−1
c=a2+12c=\frac{a^2+1}{2}c=2a2+1

对于a,b,c须满足:

a,b,ca,b,ca,b,c皆为正整数,所以a2−12\frac{a^2-1}{2}2a2−1与a2+12\frac{a^2+1}{2}2a2+1要为正整数，所以a要为奇数
a>0a>0a>0
b>=ab>=ab>=a即b=a2−12>=ab=\frac{a^2-1}{2}>=ab=2a2−1>=a得a>=2a>=2a>=2
c>bc>bc>b即a2+12>a2−12\frac{a^2+1}{2}>\frac{a^2-1}{2}2a2+1>2a2−1显然成立

综上，满足a2+12<=n\frac{a^2+1}{2}<=n2a2+1<=n的所有奇数a(a>=3)a(a>=3)a(a>=3)的个数即为所求。

代码如下：

#include<iostream>
using namespace std;
typedef long long ll;
ll n;
int main() {int t;cin >> t;while (t--) {ll cnt = 0;cin >> n;for (ll a = 3; (a * a + 1) / 2 <= n; a += 2) {cnt++;}cout << cnt << endl;}return 0;
}

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D. Pythagorean Triples

一、题目

二、输入

三、输出

四、样例

五、思路 + 代码

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