Codeforces Round #792 (Div. 1 + Div. 2) 补题

A~D

A

思路：假如不是两位数，那么一定可以找到最大的那个数，直接输出即可。如果是两位数，那就是第二位的数字

void solve()
{
	string st;
	cin >> st;
	int lens = st.size();
	if(lens == 2)
	{
		cout << st[1] << endl;
		return ;
	}
	for(int i = 0; i < st.size(); i++)
	{
		a[i] = st[i] - '0';
	} 
	sort(a, a + lens);
	cout << a[0] << endl;
}

B

思路：因为\(a < b < c\)，且由于x mod y = a, y mod z = b, z mod x = c
从前往后看，x mod y = a,那就使\(x\)比\(y\)多\(a\)就行了, 所以x = y + a, y = z + b, 按理说\(z\)应该等于x + c，但是根据题目可知\(z\)是绝对比\(x\)小的数，所以一个小数\(mod\)一个大数，结果为小数，即直接让z = c, 然后再代回去，y = c + b, x = a + b + c.

void solve()
{
	int x, y, z;
	cin >> x >> y >> z;
	cout << x + y + z << ' ' << y + z << ' ' << z << endl; 
}

C~D题不是自己想出来的，原作者是Jakon（点这里）

C

typedef long long LL;
typedef pair<int, int> PII;
const int N = 200010, M = 60, INF = 0x3f3f3f3f; 
int a[N], b[N];
bool check(int a, int b)
{
	for(int i = 0; i < n; i++)
	{
		if(v[i][a] != nv[i][b] || v[i][b] != nv[i][a]) return false;
	}
	return true;
}

int n;
vector<int> v[N], nv[N];
void solve()
{
	int m;
	cin >> n >> m;
	for(int i = 0; i < n; i++) v[i].clear();
	
	for(int i = 0; i < n; i ++)
	{
		for(int j = 0; j < m; j++)
		{
			int x;
			cin >> x;
			v[i].push_back(x);
		}
		
		nv[i] = v[i];
		sort(v[i].begin(), v[i].end());
	}
	bool flag = true;
	int pos1 = -1, pos2 = -1;
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			if(v[i][j] != nv[i][j])
			{
				if(pos1 == -1 || pos1 == j) pos1 = j;
				else if(pos2 == -1 || pos2 == j) pos2 = j;
				else flag = false;
			}
		}
	}
	
	if(!flag)
	{
		puts("-1");
		return ;
	}
	if(pos1 == -1 && pos2 == -1) puts("1 1");
	else if(check(pos1, pos2)) cout << pos1 + 1 << ' ' << pos2 + 1 << endl;
	else puts("-1");
}

D

typedef long long LL;
typedef pair<int, int> PII;
const int N = 200010, M = 60, INF = 0x3f3f3f3f; 
int a[N], b[N];
PII bs[N];
void solve()
{
	int n, k;
	cin >> n >> k;
	int c[N];
	for(int i = 1; i<= n; i++) cin>> a[i];
	LL ans = 0;
	for(int i = 1; i <= n; i++)
	{
		bs[i] = {a[i] + i, i};
	}
	sort(bs + 1, bs + 1 + n, qh);
	for(int i = 1; i <= k; i++) c[i] = bs[i].second;
	sort(c + 1, c + 1 + k);
	int cnt = 0;
	for(int i = 1; i <= n; i ++)
	{
		if(cnt < k && i == c[cnt + 1]) cnt ++;
		else ans += a[i] + cnt;
	}
	cout << ans << endl;
	
}