Educational Codeforces Round 136 (Rated for Div. 2) 补题题解 （A、B）

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本来计划着打这场的，快开始了，咳咳某人喊我打王者，没打成比赛，纯补题

A Immobile Knight

简单模拟，只有几种情况下会有孤立的点，单列出来即可

#include

using namespace std;

#define db double

const int N = 2e5 + 10;

int T;
int n, m; 

void solve(){
	cin>>n>>m;
	if(n == 1 || m == 1 || (n == 2 && m == 2)) cout<<"1 1"<<endl;\
	else if(n == 2 && m == 3) cout<<"1 2"<<endl;
	else if(n == 3 && m == 2) cout<<"2 1"<<endl;
	else if(n == 3 && m == 3) cout<<"2 2"<<endl;
	else cout<<"1 1"<<endl;
}

int main()
{
	cin>>T;
	while(T -- ){
		solve();
	}
	return 0;
}

B Array Recovery

不是难题，简单思维，目测难度应该在800~1000
简单思维：差分数组逆推原数组，原数组的每个数都大于等于0，如果某一位存在两种情况，则不合条件输出-1（偷偷吐槽一下题中也没说原数组的元素可以等于0，不然还能一发入魂）

#include

using namespace std;

#define int long long 

const int N = 2e5 + 10;

int T;
int n, m; 
int a[N], d[N];

void solve(){
	bool p = true;
	cin>>n;
	for(int i = 1; i <= n; i ++ ) cin>>d[i];
	
	a[1] = d[1];
	
	for(int i = 2; i <= n; i ++ ){
		int a1 = a[i - 1] + d[i];
		int a2 = a[i - 1] - d[i];
		if(a2 >= 0 && a1 != a2){
			p = false;
			break;
		} 
		a[i] = a1;
	}
	
	if(!p){
		cout<<"-1"<<endl;
		return ;
	} 
	
	for(int i = 1; i <= n; i ++ ) cout<<a[i]<<' ';
	
	cout<<endl;
}

signed main()
{
	cin>>T;
	while(T -- ){
		solve();
	}
	return 0;
}