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2072E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?

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思路讲解

设c为斜边，a，b为直角边

{a2+b2=c2a+b=ca2+b2=(a+b)22ab=0\begin{cases} a^2 + b^2 = c^2 \\ a + b = c \end{cases} \\ a^2+b^2=(a+b)^2 \\ 2ab=0

所以说a,b中必然有一为0。换言之，满足题设条件的充要条件是两个点在同一水平线或同一垂直线上。

AC代码

直接AC，没什么好说的

https://codeforces.com/problemset/submission/2072/309435277

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// Problem: E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?
// Contest: Codeforces - Codeforces Round 1006 (Div. 3)
// URL: https://codeforces.com/problemset/problem/2072/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(5e3)+10,INF=static_cast<ll>(5e18)+3;

ll K,T;

inline void solve(){
	cin>>K;
	vector<pll> ans;
	ll upj=1;
	for(int i=1;i<=500;++i){
		if(K==0) break;
		ll idx=0;
		for(int j=upj;j<=500+upj;++j){
			if(K-idx<0) {
				upj=j;
				break;
			}
			K-=idx;
			ans.pb({i,j});
			++idx;
		}
	}
// #ifdef LOCAL
    // cerr<<  << ans.size() << '\n';
// #endif
	cout<<ans.size()<<"\n";
	for(int i=0;i<ans.size();++i){
		cout<<ans[i].fi<<" "<<ans[i].se<<"\n";
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	return 0;
}
/*
AC
https://codeforces.com/problemset/submission/2072/309435277
*/

心路历程（WA，TLE，MLE……）