CF1523B Lord of the Values(构造+思维)

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题意:

给定一个长度为(n(2<=n<=10^{3}且为偶数))的序列,每次都可以选择两个下标(i,j)并且(i<j),进行下面两种操作之一:
(1.) (a_{i}=a_{i}+a_{j})
(2.a_{j}=a_{j}-a_{i})
最多执行(5000)次操作,并且操作后的的值不能超过(10^{18})
输出方案使得序列全部变为相反数。

思路:

对于两个数来说,按照 (1 2 1 1 2 1)的顺序执行能够将这两个数都变为自身的相反数。所以,两两的进行操作,最多操作次数为(6*n/2=3*n)次。

代码:

// Problem: B. Lord of the Values
// Contest: Codeforces - Deltix Round, Spring 2021 (open for everyone, rated, Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1523/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
const int maxn=1100;
int n,a[maxn];
int main(){
	int T;cin>>T;
	while(T--){
		cin>>n;
		for(int i=1;i<=n;i++) cin>>a[i];
		cout<<3*n<<endl;
		for(int i=1;i<=n;i+=2){
			cout<<"1 "<<i<<" "<<i+1<<endl;
			cout<<"2 "<<i<<" "<<i+1<<endl;
			cout<<"1 "<<i<<" "<<i+1<<endl;
			cout<<"1 "<<i<<" "<<i+1<<endl;
			cout<<"2 "<<i<<" "<<i+1<<endl;
			cout<<"1 "<<i<<" "<<i+1<<endl;
		}
	}
	return 0;
}


原文地址:https://www.cnblogs.com/OvOq/p/14843350.html