比赛链接：https://codeforc.es/contest/1385

pro：3.5 / 7

=====A=====

  *题意：多组数据，每组三个正整数$x, y, z$，构造$a, b, c$满足$x = max(a, b)$, $y = max(a, c)$, $z= max(b, c)$
  *题解：签到题，分类特判即可
<hidden ac代码>
<code c++>
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 2e5 + 5;
const double pi = acos(-1);
const int mod = 998244353;

int x, y, z, a, b, c;
int t;

int main(){
	cin >> t;
	while(t--){
		cin >> x >> y >> z;
		int a = (x == y) + (x == z) + (y == z);
		if(!a){
			puts("NO");
			continue;
		}
		else if(a == 3){
			puts("YES");
			cout << x << ' ' << x << ' ' << x << endl;
			continue; 
		}
		else{
			if((x == y && z > x) || (x == z && y > x) || (y == z && x > y)) puts("NO");
			else{
				puts("YES");
				if(x == y) cout << x << ' ' << z << ' ' << z << endl;
				else if(x == z) cout << x << ' ' << y << ' ' << y << endl;
				else cout << y << ' ' << x << ' ' << x << endl;
			}
		}
	}
} 
</code>
</hidden>

=====B=====

  *题意：给定多组有且仅有两次重复的数字序列，在不影响相对顺序的前提下去重
  *题解：签到题，打标记跳输出即可
<hidden ac代码>
<code c++>
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 2e5 + 5;
const double pi = acos(-1);
const int mod = 998244353;

int t;
int n;
int a[105];

int main(){
	cin >> t;
	while(t--){
		cin >> n;
		int tmp;
		memset(a, 0, sizeof(a));
		for(int i = 1; i <= 2 * n; ++i){
			cin >> tmp;
			if(a[tmp]) continue;
			cout << tmp << ' ';
			a[tmp] = 1;
		}
		puts("");
	}
} 
</code>
</hidden>

=====C=====

  *题意：定义“good”序列：每次从首位或末位拿出元素，组成的新序列单调不降。多组数据，给定序列，询问删去从首位开始的至少多少元素，可以使序列成为“good”序列？
  *题解：所谓“good”序列就是从首末开始到某一位置都单调不降的序列，只需倒序遍历，先找不降序列，由此后找不升序列，统计剩余元素个数输出即可
<hidden ac代码>
<code c++>
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 2e5 + 5;
const double pi = acos(-1);
const int mod = 998244353;

int t;
int n;
int a[maxn];

int main(){
	cin >> t;
	while(t--){
		cin >> n;
		for(int i = 1; i <= n; ++i) cin >> a[i];
		if(n == 1){
			cout << 0 << endl;
			continue;
		}
		int p = -1, q = -1;
		for(int i = n; i > 1; --i){
			if(a[i-1] < a[i]){
				p = i;
				break;
			}
		}
		if(p == -1){
			cout << 0 << endl;
			continue;
		}
		for(int i = p; i > 1; --i){
			if(a[i-1] > a[i]){
				q = i;
				break;
			}
		}
		if(q == -1) cout << 0 << endl;
		else cout << q - 1 << endl;
	}
} 
</code>
</hidden>

=====D=====

  *见本周推荐