Codeforces Round 1006 (Div. 3) E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?

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

题目链接👈

题目描述🥰

题目思路😀

这个题目首先我们可以关注到当p(i,j)=d(i,j),这种情况只会存在两个坐标的x坐标相同或是y相同，所以当x个员工站成一条线，我们就会有x*(x-1)/2个配对。

所以题目的解法就很明显了，我们只需要关注有多少个小于k的数m（m=x*(x-1)/2，x>=0&&x<m)能够凑成k，然后构造答案即可.

但是凑成k这个代码如果书写？解释如下

//迭代计算由几个值构成k
vector<int> dev(int k)
{
	vector<int>ans;
	while(k>0)
	{
		int m=(sqrt(1+k*8)+1)/2;
		while(m*(m-1)/2>k)m--;
		ans.push_back(m);
		k-=(m-1)*m/2;
	}
	return ans;
}

AC代码🧠

// 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/contest/2072/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define dev1(a) cout << #a << '=' << a << endl;
#define dev2(a, b) cout << #a << " = " << a << "  " << #b << " = " << b << endl;
#define dev3(a, b, c) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << endl;
#define dev4(a, b, c, d) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << endl;
#define dev5(a, b, c, d, e) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << "  " << #e << " = " << e << endl;
#define vec(a)                         \
    for (int i = 0; i < a.size(); i++) \
        cout << a[i] << ' ';           \
    cout << endl;
#define darr(a, _i, _n)               \
    cout << #a << ':';                \
    for (int ij = _i; ij <= _n; ij++) \
        cout << a[ij] << ' ';         \
    cout << endl;           
#define cin(a,n)           \
     for(int i=0;i<n;i++) \
      cin>>a[i];          
#define endl "\n"
#define pow pim
int pim(int a,int k)
{
    int res=1;
    if(a==0)return 0;
    while(k) 
    {
        if(k&1)res=(int)res*a;
        k>>=1;
        a=(int)a*a;
    }
    return res;
}
#define fi first
#define se second
#define caseT \
    int T;    \
    cin >> T; \
    while (T--)
// #define int long long
// #define int __int128

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 99999999;

// const int N = ;
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
 
int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
//迭代计算由几个值构成k
vector<int> dev(int k)
{
	vector<int>ans;
	while(k>0)
	{
		int m=(sqrt(1+k*8)+1)/2;
		while(m*(m-1)/2>k)m--;
		ans.push_back(m);
		k-=(m-1)*m/2;
	}
	return ans;
}
void solve()
{
	int k;
	cin>>k;
	if (k == 0) {
        cout << 0<<endl;
        return ;
    }
	vector<int>a=dev(k);
	int sum=0;
	for(auto x:a)sum+=x;
	cout<<sum<<endl;
	int x=0,y=0;
	for(auto t:a)
	{
		for(int i=0;i<t;i++)
		{
			cout<<x<<" "<<y+i<<endl;
		}
		y+=t;
		x++;
	}
}

signed main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	caseT
    solve();
    return 0;
}
/*

*/