2022.5.12
Codeforces Round #790 (Div. 4)
A - Lucky?
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int a[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
string s;
cin >> s;
int ans = (s[0] - '0') + (s[1] - '0') + (s[2] - '0');
int ans1 = (s[3] - '0') + (s[4] - '0') + (s[5] - '0');
if(ans==ans1)
cout << "yes\n";
else
cout << "no\n";
}
return 0;
}
B - Equal Candies
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int a[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
int n;
cin>>n;
int mmin = INF;
for (int i = 1; i <= n ;i++)
{
cin >> a[i];
mmin = min(a[i], mmin);
}
ll sum = 0;
for (int i = 1;i<=n;i++)
{
sum += a[i] - mmin;
}
cout << sum << '\n';
}
return 0;
}
C - Most Similar Words
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
string s[60];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
int n,m;
cin>>n>>m;
for (int i = 1; i <= n;i++)
{
cin >> s[i];
}
int ans = INF;
for (int i = 1; i < n;i++)
{
for (int j = i + 1; j <= n;j++)
{
int res = 0;
for (int k = 0; k < m;k++)
{
res += abs(s[i][k] - s[j][k]);
}
ans = min(ans, res);
}
}
cout << ans << '\n';
}
return 0;
}
D - X-Sum
由于n很小直接暴力搜即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=200+10,INF=1e9;
int a[N][N],n,m;
int dx[] = {-1, 1, 1, -1}, dy[] = {1, 1, -1, -1};
bool check(int x,int y)
{
return x >= 1 && x <= n && y >= 1 && y <= m;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
ll ans = -1;
cin >> n >> m ;
for (int i = 1;i<=n;i++)
{
for (int j = 1; j <= m;j++)
{
cin >> a[i][j];
}
}
for (int i = 1; i <= n;i++)
{
for (int j = 1; j <= m;j++)
{
ll res = 0;
int cnt = 0;
for (int k = 0; k < 4;k++)
{
int x = i, y = j;
if(x==i&&y==j)
{
cnt++;
}
while(check(x,y))
{
res += a[x][y];
x += dx[k], y += dy[k];
}
}
cnt--;
res -= cnt * a[i][j];
ans = max(ans, res);
}
}
cout << ans << '\n';
}
return 0;
}
E - Eating Queries
前缀和+二分
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=2e5+10,INF=1e9;
int a[N];
ll sum[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
int n,m;
cin >> n >> m;
memset(sum,0, sizeof sum);
for (int i = 1; i <= n;i++)
{
cin >> a[i];
}
sort(a + 1, a + 1 + n);
for (int i = n; i >= 1; i--)
{
sum[i]=sum[i+1]+a[i];
}
reverse(sum + 1, sum + 1 + n);
while (m--)
{
ll x;
cin >> x;
if (x > sum[n])
cout << "-1\n";
else
{
int cnt = lower_bound(sum + 1, sum + 1 + n, x)-(sum);
cout << cnt << '\n';
}
}
}
return 0;
}
F - Longest Strike
去重+map记录，边循环边判断，更新最大值
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=2e5+10,INF=1e9;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--)
{
map<int, int> vis;
int n, m;
cin >> n >> m ;
int mmax = 0, maxpos = 0;
vector<int> a;
for (int i = 1; i <= n;i++)
{
int x;
cin >> x;
if(!vis[x])
a.push_back(x);
vis[x]++;
if(vis[x]>mmax)
{
mmax = vis[x];
maxpos = x;
}
}
if(mmax<m)
{
cout << "-1\n";
}
else
{
sort(a.begin(), a.end());
int res = -1;
pii ans = {0, 0};
int pos = 0;
int sz = a.size();
for (int i = 1; i < sz;i++)
{
if(a[i]==a[i-1]+1&&vis[a[i-1]]>=m&&vis[a[i]]>=m)
{
int len = a[i] - a[pos];
if(len>res)
{
res = len;
ans = {a[pos], a[i]};
}
}
else
pos = i;
}
if(ans.first==0&&ans.second==0)
{
ans = {maxpos, maxpos} ;
}
cout << ans.first << ' ' << ans.second << '\n';
}
}
return 0;
}
G - White-Black Balanced Subtrees
简单DFS,每次返回子树白色和黑色点的个数，如果相同cnt++，最后判断以1为根节点的数颜色是否相等即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=4e3+10,INF=1e9;
vector<int> p[N];
map<int, char> mp;
ll cnt = 0;
bool vis[N];
pii dfs(int x)
{
pii sum;
if(mp[x]=='W')
sum = {1, 0};
else
sum = {0, 1};
if(!x)
return sum;
for (int i = 0, j = p[x].size(); i < j;i++)
{
int son = p[x][i];
if(!vis[son])
{
vis[son] = 1;
pii res = dfs(son);
if(res.first==res.second)
cnt++;
sum = {sum.first + res.first, sum.second + res.second};
}
}
return sum;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cnt = 0;
int n;
cin>>n;
for (int i = 0; i <= n;i++)
{
p[i].clear();
vis[i] = 0;
}
for (int i = 2; i <= n;i++)
{
int x;
cin>>x;
p[x].push_back(i);
}
string s;
cin >> s;
for (int i = 0; i < n;i++)
{
mp[i + 1] = s[i];
}
pii res=dfs(1);
if(res.first==res.second)
cnt++;
cout << cnt << '\n';
}
return 0;
}
H1 - Maximum Crossings (Easy Version)
求逆序对加了个等于
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e3+10,INF=1e9;
int a[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
int n;
cin >> n;
for (int i = 1; i <= n;i++)
{
cin >> a[i];
}
int cnt = 0;
for (int i = 2; i <= n;i++)
{
for (int j = 1; j < i;j++)
{
if(a[j]>=a[i])
cnt++;
}
}
cout << cnt << '\n';
}
return 0;
}
H2 - Maximum Crossings (Hard Version)
归并排序求逆序对，也可以树状数组
#include<iostream>
using namespace std;
const int maxn=1e6+10;
int a[maxn],tmp[maxn];
long long cnt;
void mergesort(int l,int r)
{
if(l>=r) return ;
int mid=l+r>>1;
mergesort(l,mid);
mergesort(mid+1,r);
int k=0,i=l,j=mid+1;
while(i<=mid&&j<=r)
{
if(a[i]<a[j]) tmp[k++]=a[i++];
else
{
tmp[k++]=a[j++];
cnt+=mid-i+1;
}
}
while(i<=mid) tmp[k++]=a[i++];
while(j<=r) tmp[k++]=a[j++];
for(int i=l,j=0;i<=r;j++,i++) a[i]=tmp[j];
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int n;
cnt = 0;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
mergesort(1,n);
cout << cnt << '\n';
}
return 0;
}
树状数组
#include<bits/stdc++.h>
const int N=200005;
using namespace std;
typedef long long ll;
ll a[N],d[N],tree[N],n;
int lowbit(int x)
{
return x&-x;
}
void add(int x)
{
while(x<=n)
{
tree[x]++;
x+=lowbit(x);
}
}
ll sum(int x)
{
ll res=0;
while(x>=1)
{
res+=tree[x];
x-=lowbit(x);
}
return res;
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
ll ans=0;
scanf("%lld",&n);
for (int i = 0; i <= n;i++)
tree[i] = 0;
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]);
for(int i=1;i<=n;i++)
{
add(a[i]);
ans+=i-1-sum(a[i]-1);
}
printf("%lld\n",ans);
}
return 0;
}
