2022ICPC济南 - E K M

目录

• E. Identical Parity
• K. Stack Sort
• M. Best Carry Player

2022 International Collegiate Programming Contest, Jinan Site
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E. Identical Parity

无论 k 怎么给定，k 个数里面奇数个数要么和偶数相等，要么奇数比偶数多一个（因为总体的奇数个数可能比偶数个数多一个），此时再利用余数去补足即可

少点感觉是怎么回事。。。

void solve(){
	int n, k;
	cin >> n >> k;
	if(n == 1) cout << "Yes\n";
	else if(k % 2 == 0) cout << "Yes\n";
	else{
		int cnta = (n + 1) / 2, cntb = n / 2;
		int numa = k / 2 + 1, numb = k / 2;
		int x = n / k, r = n % k;
		int lefta = cnta - x * numa, leftb = cntb - x * numb;
		if(lefta >= 0 && leftb >= 0 && lefta + leftb == r){
			if(lefta <= numa && leftb <= numb){
				cout << "Yes\n";
				return ;
			}
		}
		cout << "No\n";
	}
	return ;
}

K. Stack Sort

从前往后遍历标记每个数，如果当前数 + 1 出现过则跟到前面的栈里面，没出现过就新开一个栈

void solve(){
	int n, a, ans = 0;
	cin >> n;
	vector<bool> vis(n + 2, false);
	for(int i = 0; i < n; ++ i){
		cin >> a;
		if(!vis[a + 1]) ++ ans;
		vis[a] = true;
	}
	cout << ans << '\n';
	return ;
}

M. Best Carry Player

模拟进位即可，记得数组开大点就行

void solve(){
	int n;
	cin >> n;
	vector<int> d(100, 0);
	for(int i = 0; i < n; ++ i){
		int a, c = 0;
		cin >> a;
		while(a){
			d[c] += a % 10;
			a /= 10;
			++ c;
		}
	}
	ll ans = 0;
	for(int i = 0; i < 99; ++ i){
		ans += d[i] / 10;
		d[i + 1] += d[i] / 10;
	}
	cout << ans << '\n';
	return ;
}