Group Programming ladder match L1-056 ~ L1-060

L1-056

Thinking:

Here emissions calculated relative to a sequence

of code:

#include<bits/stdc++.h>

using namespace std;

typedef pair<int, string> P;

int main() {	
	int n, ans = 0;
	cin >> n;
	vector<P> v;
	for(int i = 0; i < n; i++){
		string s;
		int num;
		cin >> s >> num;
		v.push_back(P(num, s));
		ans += num;
	}
	ans = int(1.0 * ans / n / 2);
	cout << ans << ' ';
	for(P & p : v) p.first = abs(p.first - ans);
	sort(v.begin(), v.end());
	cout << v[0].second;
	return 0;
}

L1-057

Thinking:

really I do not want to play the phrase deceptive words

of code:

#include<bits/stdc++.h>

using namespace std;

int main() {	
	puts("PTA shi3 wo3 jing1 shen2 huan4 fa1 !");
	return 0;
}

L1-058

Ideas:

This title Do not use a space to separate them, one by one loop it honestly

Code:

#include<bits/stdc++.h>

using namespace std;

int main() {	
	string s;
	getline(cin, s);
	for(int i = 0; i < s.length(); i++){
		if(s[i] != '6') { putchar(s[i]); continue; }
		for(int j = i; j < s.length(); j++){
			if(j + 1 == s.length() || s[j + 1] != '6'){
				int len = j - i + 1;
				if(len > 9) { cout << 27; i = j; break; }
				else if(len > 3) { cout << 9; i	= j; break; }
				else { cout << s.substr(i, len); i = j; break; }
			}
		}
	}
	return 0;
}

L1-059

Thinking:

This is the question a little nausea , analog bar

codes:

#include<bits/stdc++.h>

using namespace std;

bool ok(string & s){
	string cmp = "ong";
	int x = s.find(','), y = s.find('.');
	if(x < 3) return false;
	if(s.substr(x - 3, 3) != cmp || s.substr(y - 3, 3) != cmp) return false;
	reverse(s.begin(), s.end());
	for(int i = 0; i < 3; i++){
		int pos = s.find(' ');
		s = s.substr(pos + 1);	
	}
	reverse(s.begin(), s.end());
	return true;
}

int main() {	
	int n;
	cin >> n;
	getchar();
	while(n--){
		string s;
		getline(cin, s);
		if(ok(s)) cout << s << " qiao ben zhong.\n";
		else cout << "Skipped\n";	
	}
	return 0;
}

L1-060

Ideas:

a large area to reduce the area

of code:

#include<bits/stdc++.h>

using namespace std;

#define s(x, y) ((x) * (y) / 2)
int main() {	
	int x, y;
	cin >> x >> y;
	cout << 5000 - s(x, y) - (100 -x) * y - s(100 - x, 100 -y);
	return 0;
}