A
特判一下前面是10，后面>2即可。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve()
{
string s;
cin >> s;
if (s[0] == '1' && s[1] == '0' && (s[2] >= '2' || s[2] == '1' && s[3] >= '0' && s[3] <= '9'))
{
cout << "YES" << '\n';
}
else
{
cout << "NO" << '\n';
}
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
B
开个map记录即可。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve()
{
map<int,int>mp;
int n;
cin>>n;
bool flag=0;
for(int i=1;i<=n;i++)
{
int x;
cin>>x;
if(i==1)
{
mp[x]=1;
continue;
}
if(!mp[x-1]&&!mp[x+1])
{
flag=1;
}
mp[x]=1;
}
if(!flag)
{
cout<<"YES"<<'\n';
}
else
{
cout<<"NO"<<'\n';
}
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
C
数字到字母做一个映射。
字母到数字做一个映射。
还是map。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve()
{
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
int m;
cin >> m;
for (int i = 1; i <= m; i++)
{
string s;
cin >> s;
map<char, int> mp;
bool flag=0;
if (s.size() != n)
{
cout << "NO" << '\n';
continue;
}
for (int j = 0; j < s.size(); j++)
{
if (mp.count(s[j]) && mp[s[j]] != a[j + 1])
{
flag=1;
cout << "NO" << '\n';
break;
}
else
{
mp[s[j]] = a[j + 1];
}
}
if(flag)
{
continue;
}
map<int,char>mp1;
for (int j = 0; j < s.size(); j++)
{
if (mp1.count(a[j+1]) && mp1[a[j+1]] != s[j])
{
flag=1;
cout << "NO" << '\n';
break;
}
else
{
mp1[a[j+1]] = s[j];
}
}
if(flag)
{
continue;
}
cout << "YES" << '\n';
}
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
D
赛时太nt了，猜了个一坨出来。
只需要每一个L去匹配最远的R即可。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve()
{
int n;
cin >> n;
vector<ll> a(2 * n);
vector<char> b(2 * n);
vector<ll> sum(2 * n);
for (int i = 1; i <= n; i++)
{
cin >> a[i];
sum[i] = sum[i - 1] + a[i];
}
string s;
cin >> s;
s = " " + s;
vector<int>p(n+1);
for(int i=1;i<=n;i++)
{
if(s[i]=='R')
{
p.push_back(i);
}
}
ll ans=0;
for(int i=1;i<=n;i++)
{
if(s[i]=='L')
{
if(p.empty())
{
break;
}
int x=p.back();
p.pop_back();
if(x<i)
{
break;
}
ans+=sum[x]-sum[i-1];
}
}
cout<<ans<<'\n';
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
E
预处理出每个格子被覆盖的次数，直接用前缀和和差分，然后用优先队列即可。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve()
{
int n, m, k;
cin >> n >> m >> k;
ll w;
cin >> w;
vector<ll> a(w + 1);
for (int i = 1; i <= w; i++)
{
cin >> a[i];
}
sort(a.begin() + 1, a.end());
vector<vector<ll>> sum(n + k + 1, vector<ll>(m + k + 1));
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
int x2 = i + k - 1;
int y2 = j + k - 1;
if (x2 > n || y2 > m)
{
continue;
}
sum[i][j]++;
sum[i][y2 + 1]--;
sum[x2 + 1][j]--;
sum[x2 + 1][y2 + 1]++;
}
}
priority_queue<ll> q;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
q.push(sum[i][j]);
}
}
ll ans = 0;
for (int i = w; i >= 1; i--)
{
if (q.empty())
{
break;
}
ans += q.top() * a[i];
q.pop();
}
cout << ans << '\n';
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
F
预处理每个矩形得分对应的操作次数，然后直接dp。
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| #include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1e18;
void solve()
{
int n, k;
cin >> n >> k;
vector<ll> dp(k + 1, INF);
dp[0] = 0;
for (int i = 0; i < n; i++)
{
int a, b;
cin >> a >> b;
vector<ll> me(k + 1, INF);
me[0] = 0;
int cnt = 0, cost = 0;
while (cnt < k && (a > 0 || b > 0))
{
if (a < b)
{
swap(a, b);
}
cnt += 1;
cost += b;
a -= 1;
me[cnt] = cost;
}
for (int j = k - 1; j >= 0; j--)
{
for (int p = 1; p <= k - j; p++)
{
dp[j + p] = min(dp[j + p], dp[j] + me[p]);
}
}
}
cout << (dp[k] == INF ? -1 : dp[k]) << '\n';
return;
}
int main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
|
