The Right-angled Triangles
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The Right-angled Triangles
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https://ac.nowcoder.com/acm/contest/338/K
題解:竟然暴力就行???
枚舉一條直角邊的長(zhǎng)度,印證c*c-(枚舉的直角邊的平方)是不是應(yīng)該平方數(shù);為了減少時(shí)間枚舉到c*c/2就行了
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q,ans; bool a[N]; char str; bool sloved(ll c){ll c2=c*c;for(int i=1;i*i<=c2/2;i++){q=sqrt(c2-i*i);if(q*q==c2-i*i)return 1;}return 0; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifwhile(~scanf("%d",&t))while(t--){scanf("%d",&n);if(sloved(n)){cout << "Yes" << endl;}else{cout << "No" << endl;}}//cout << "Hello world!" << endl;return 0; }?
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