https://codeforces.com/contest/1487/problem/D
题目大意
给出一个上界 $n \ (\le 10^9)$,问有多少正整数三元组 $(a, b, c)$ 满足 $(1 \le a \le b \le c \le n)$, $a^2 + b^2 = c^2$ 且满足 $c = a^2 - b$。
简要题解
由 $a^2 + b^2 = c^2$ 和 $c = a^2 - b$ 都包含 $a^2$ 则我们可以将其约去,易得 $c = b + 1$。
注意到 $a^2 = b + c$ 则 $a \le \sqrt{2n}$。于是枚举 $a$ 进而通过 $a^2 = b + c$ 和 $c = b + 1$ 得到 $b, c$ 并验证即可。
复杂度
$T$:$O(sqrt(n))$
$S$:$O(1)$
代码实现
#include <bits/stdc++.h>
using namespace std;
int io_=[](){ ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();
using LL = long long;
using ULL = unsigned long long;
using LD = long double;
using PII = pair<int, int>;
using VI = vector<int>;
using MII = map<int, int>;
template<typename T> void cmin(T &x,const T &y) { if(y<x) x=y; }
template<typename T> void cmax(T &x,const T &y) { if(x<y) x=y; }
template<typename T> bool ckmin(T &x,const T &y) {
return y<x ? (x=y, true) : false; }
template<typename T> bool ckmax(T &x,const T &y) {
return x<y ? (x=y, true) : false; }
template<typename T> void cmin(T &x,T &y,const T &z) {// x<=y<=z
if(z<x) { y=x; x=z; } else if(z<y) y=z; }
template<typename T> void cmax(T &x,T &y,const T &z) {// x>=y>=z
if(x<z) { y=x; x=z; } else if(y<z) y=z; }
// mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
// mt19937_64 rnd_64(chrono::system_clock::now().time_since_epoch().count());
/*
---------1---------2---------3---------4---------5---------6---------7---------
1234567890123456789012345678901234567890123456789012345678901234567890123456789
*/
/*
1 <= a <= b <= c <= n
a^2 + b^2 = c^2
a^2 - b = c
b + c = c^2 - b^2 = (c + b) (c - b)
-> c - b = 1
*/
void solve() {
LL n; cin >> n;
int ans = 0;
for (LL a = 1; a * a <= n * 2; a += 2) {
LL a2 = a * a;
LL b = a2 / 2, c = b + 1;
if (a <= n && 1 <= b && b <= n && c <= n && a2 + b * b == c * c) {
// cerr << a << ' ' << b << ' ' << c << endl;
ans++;
}
}
cout << ans << '\n';
}
int main() {
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
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