团体程序设计天梯赛 L1-016~L1-020

L1-016

思路：

按题意操作即可

代码：

#include<bits/stdc++.h>

using namespace std;

const int p = 17;
const int mod = 11;
int w[] = {7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2};
int m[] = {1, 0, 11, 9, 8, 7, 6, 5, 4, 3, 2};

bool ok(string& s) {
	int ans = 0;
	for(int i = 0; i < p; i++) {
		int now = s[i] - '0';
		ans += now * w[i];
	}
	int mm = s[p] == 'X' ? 11 : s[p] - '0';
	return mm == m[ans % mod];
}
int main() {	
	int n;
	cin >> n;
	vector<string> v;
	for(int i = 0; i < n; i++) {
		string s;
		cin >> s;
		if(!ok(s)) v.push_back(s);	
	}
	if(v.size() == 0) puts("All passed");
	else for(string& s : v) puts(s.c_str());
	return 0;
}

L1-017

思路：

按题意操作即可，注意输出精度

代码：

#include<bits/stdc++.h>

using namespace std;

int main(){	
	double a = 0, b, c = 1, d = 1;
	string s;
	cin >> s;
	b = s.length();
	for(char& c : s) if(c == '2') a += 1;
	if(((s[b - 1] - '0') & 1) == 0) d = 2;
	if(s[0] == '-') c = 1.5, --b;
	printf("%.2f%%", 100.0 * a / b * c * d);
	return 0;
}

L1-018

思路：

按题意操作即可

代码：

#include<bits/stdc++.h>

using namespace std;

int main(){	
	int a, b;
	cin >> a;
	getchar();
	cin >> b;
	if(a * 60 + b <= 12*60) printf("Only %02d:%02d.  Too early to Dang.", a, b);
	else{
		int num = a - 12 + !!b;
		while(num--) cout << "Dang";	
	}
	return 0;
}

L1-019

思路：

按题意进行模拟

代码：

#include<bits/stdc++.h>

using namespace std;

int main(){	
	int x, y, n;
	cin >> x >> y >> n;
	int xx = 0, yy = 0;
	while(n--){
		int a, b, c, d;
		cin >> a >> b >> c >>d;
		bool _x = (b == a + c);
		bool _y = (d == a + c);
		if(_x == 1 && _y == 0) 	++xx;
		else if(_x == 0 && _y == 1) ++yy;
		if(xx > x) cout << "A
" << yy, exit(0);
		if(yy > y) cout << "B
" << xx, exit(0);
	}
}

L1-020

思路：

开个数组映射一下即可；

代码：

#include<bits/stdc++.h>

using namespace std;

const int maxn = 123456;
bool tag[maxn];

int main(){	
	int n, k, m, x;
	cin >> n;
	while(n--){
		cin >> k;
		for(int i = 0; i < k; i++){
			cin >> x;
			if(k > 1) tag[x] = true;
		}
	}
	cin >> m;
	vector<int> rs;
	while(m--){
		cin >> x;
		if(tag[x] == false) rs.push_back(x), tag[x] = true;
	}
	if(rs.size()){
		printf("%05d", rs[0]);
		for(int i = 1; i < rs.size(); i++) printf(" %05d", rs[i]); 	
	}
	else puts("No one is handsome");
	return 0;
}