A. Extremely Round
当n为3位数时,例如\(n=120\),满足题目要求的情况有
以上19种情况,一位和二位去满各有九种情况,三位只能取一种情况,所以共19种
code
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
void solve()
{
int n;
cin >> n;
int res = 0;
while(n)
{
res += min(n,9);
n /= 10;
}
cout << res << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
cin >> t;
while(t--)
solve();
return 0;
}
B. Notepad#
只要有两个相同的两位子串,就说明可以用小于n次操作将字符串输入
code
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
bool vis[30][30];
void solve()
{
int n;
cin >> n;
string s;
cin >> s;
memset(vis,0,sizeof vis);
for(int i = 1;i < n;++i)
{
int x = s[i - 1] - 'a',y = s[i] - 'a';
if(vis[x][y])
{
puts("YES");
return;
}
vis[x][y] = true;
if(x == y && ((i < n - 1) && y == s[i + 1] - 'a'))
++i;
}
puts("NO");
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
cin >> t;
while(t--)
solve();
return 0;
}
C. Hamiltonian Wall
这里的话,我们将每一纵坐标标记为三种状态,\(c_{1,j}=B\)时为状态1,\(c_{2,j}=B\)时为状态2,均为\(B\)时为状态3
以下两种情况不能一笔画完:
- 状态1(2), 偶数个状态3,状态2(1)前后状态不同
- 状态1(2), 偶数个状态3,状态1(2)前后状态相同
code
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define vint vector<int>
void solve()
{
int n;
cin >> n;
string a, b;
cin >> a >> b;
vint num(n,0);
int idx = -1;
for (int i = 0; i < n; ++i)
{
if (a[i] == 'B')
num[i] += 1;
if (b[i] == 'B')
num[i] += 2;
if(num[i] != 3)
{
if(idx == -1)
idx = i;
else
{
if((i - idx & 1) == abs(num[i] - num[idx]))
{
puts("NO");
return;
}
idx = i;
}
}
}
puts("YES");
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
cin >> t;
while (t--)
solve();
return 0;
}
D. Lucky Chains
有且仅有两者差值为1的时候,链可以无限长
然后通过差值寻找答案即可,当差值为k的时候,长度一定是k的因数
code
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define vint vector<int>
const int N = 1e7;
int vis[N + 1];
vint prime;
void solve()
{
int x, y;
cin >> x >> y;
int d = y - x;
if (d == 1)
{
cout << -1 << endl;
return;
}
int res = inf;
while(d > 1)
{
int p = vis[d];
d /= p;
res = min(res,(p - x % p) % p);
}
cout << res << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
for (int i = 2; i <= N; ++i)
{
if (!vis[i])
prime.pb(i),vis[i] = i;
for (int j : prime)
{
if (j * i > N)
break;
vis[j * i] = j;
if (i % j == 0)
break;
}
}
int t = 1;
cin >> t;
while (t--)
solve();
return 0;
}
标签:Educational,Rated,return,int,Codeforces,cin,vis,num,solve
From: https://www.cnblogs.com/KicamonIce/p/16979628.html