Codeforces Round #720 (Div. 2) ABCDE题解

Posted 2021-07-19 quinn18

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A - Nastia and nearly Good Numbers

题目链接
构造
题意：给你 $A, B$ 输出 $x, y, z$ , 满足 $x + y = z$ ，两个个只能被A整除，一个能被A*B整除

#include<bits/stdc++.h>
using namespace std;
const int N=15;
const int M=1e9+7;
#define ll long long
ll t, n, l, r, sum, cnt, q, a[N];
string s;
int main(){
	cin>>t;
	while(t--){
		cin>>l>>r;
		if(r==1) cout<<"NO"<<endl;
		else {
            cout<<"YES"<<endl;
            cout<<(r-1)*l<<" "<<(r+1)*l<<" "<<(2*r)*l<<endl;
		}
	}
	return 0;
}

B - Ordinary Numbers

tmlj
构造
题意：最多 $n$ 次修改随便两个数，使数列中相邻两个数gcd为1 修改条件是你选中的两个数 $m i n (a i, a j) = = m i n (x, y), a i = x, a j = y 。$

找最小的数往两边扩展开来一直 $+ 1 + 1 + 1$ 毕竟相邻的数字肯定是互质的
就一直拿这个最小的数拿来和别人交换

#include<bits/stdc++.h>
using namespace std;
const int N=100005;
const int M=1e9+7;
#define ll long long
ll t, n, l, r, sum, cnt, q, a[N], minn;
string s;
int main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>t;
	while(t--){
		cin>>n;
		minn=1e18;;
		for(int i=1; i<=n; i++) {
            cin>>a[i];
            if(a[i]<minn) {
                minn=a[i];
                q=i;//找最小值
            }
		}
		cout<<n-1<<endl;
		for(int i=1; i<=n; i++) {
            if(i==q) continue;
            if(i<q)cout<<i<<" "<<q<<" "<<q-i+minn<<" "<<minn<<endl;
            else  cout<<i<<" "<<q<<" "<<i-q+minn<<" "<<minn<<endl;
		}
	}
	return 0;
}