Codeforces Round 975 (Div. 2)题解记录（A，B，C，E）
这次头有点晕，A题2分钟秒了，B题想法对的，但是头晕一直没把式子写对，纯纯掉分了
A. Max Plus Size
暴力模拟
#include<iostream>
#include<map>
#include<set>
#include<vector>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<bitset>
#include<stack>
#include<algorithm>
#include<deque>
#include<random>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans *= x;
x *= x;
y >>= 1;
}
return ans;
}
void fio()
{
ios::sync_with_stdio();
cin.tie(0);
cout.tie(0);
}
ll a[250000];
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n;
cin >> n;
for (ll i = 1; i <= n; i++)cin >> a[i];
ll k1 = 0, k2 = 0;
ll ans = 0, cnt = 0;
for (ll i = 1; i <= n; i++)
{
if (i % 2 == 0)
{
ans++; k1 = max(k1, a[i]);
}
else
cnt++, k2 = max(k2, a[i]);
}
cout << max(ans + k1, cnt + k2) << endl;
}
}
B. All Pairs Segments
不难，看懂题意就会了，求正好被扫过某个次数的给定区间数字的个数
#include<iostream>
#include<map>
#include<set>
#include<vector>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<bitset>
#include<stack>
#include<algorithm>
#include<deque>
#include<random>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans *= x;
x *= x;
y >>= 1;
}
return ans;
}
void fio()
{
ios::sync_with_stdio();
cin.tie(0);
cout.tie(0);
}
ll a[250000];
ll b[250000];
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n, q;
cin >> n >> q;
for (ll i = 1; i <= n; i++)cin >> a[i];
map <ll, ll>f;
ll ans = 0;
for (ll i = 1; i <= n ; i++)
{
f[n - i + (i - 1) * (n - i + 1)]++;
if (i != n)
{
f[i * (n - i)] += a[i+1] - a[i]-1;
}
}
for (ll i = 1; i <= q; i++)
{
cin >> b[i];
}
for (ll i = 1; i <= q; i++)
{
cout << f[b[i]] << " ";
}
cout << endl;
}
}
C. Cards Partition
赛后补的，感觉本题其实有个隐藏结论。对于符合的长度，个数最多的数都是有参与组成的，所以直接枚举长度看是否出现一个最大使这个结论成立的
成立即break,感觉用sum比的时候就好似形成了一种dp，分布最均匀化。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[250000];
int main()
{
ll t;
cin>>t;
while(t--)
{
ll n,m;
cin>>n>>m;
ll op=0,sum=0;
for(ll i=1;i<=n;i++)cin>>a[i],op=max(op,a[i]),sum+=a[i];
sort(a+1,a+1+n);
ll ans=1;
for(ll i=n;i>=1;i--)
{
ll x=i*op;
if(x>sum+m)
continue;
else
{
if(sum>=x)
{
ll j=sum%i;
if(m>=i-j||j==0)
{
ans=i;
break;
}
else
continue;
}
else
{
ans=i;
break;
}
}
}
cout<<ans<<endl;
}
}
E. Tree Pruning
直接暴力把所有长度先记录下来，
暴力所有留下长度的可能性，用\(dfs\)往上爬，通过看儿子是否全被标记再考虑是否往上爬，这样子就可以把短的边去掉了
\(Onlog(n)\)
#include<iostream>
#include<map>
#include<set>
#include<vector>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<bitset>
#include<stack>
#include<algorithm>
#include<deque>
#include<random>
using namespace std;
typedef long long ll;
ll ko = 0;
ll faa[500005];
void fio()
{
ios::sync_with_stdio();
cin.tie(0);
cout.tie(0);
}
ll sz[500025];
vector<ll>g[500050];
set<ll>q[500050];
set <ll>f;
ll vis[550000];
ll dfs1(ll x)
{
ll cnt = 0;
if (x == 1)
return 0;
if (vis[x]==g[x].size()-1)
{
vis[faa[x]]++;
return dfs1(faa[x])+1;
}
else
return 0;
}
void dfs(ll x,ll fa,ll sd)
{
sz[x] = 1;
q[sd].insert(x);
f.insert(sd);
faa[x] = fa;
for (auto j : g[x])
{
if (j == fa)
continue;
dfs(j, x,sd+1);
sz[x] += sz[j];
}
return;
}
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n;
cin >> n;
f.clear();
for (ll i = 1; i <= n; i++)
{
faa[i] = 0;
g[i].clear();
sz[i] = 0;
vis[i] = 0;
q[i].clear();
}
for (ll i = 1; i <= n - 1; i++)
{
ll l, r;
cin >> l >> r;
g[r].push_back(l);
g[l].push_back(r);
}
dfs(1, 0, 0);
for (ll i = 1; i <= n; i++)vis[i] = 0;
ll ans = 0;
ll u = 9999999999;
ll ou = 0;
ko = 0;
for (auto j : f)
{
ans = 0;
if (j != 0)
{
ou = ko;
for (auto k : q[j])
{
ans += sz[k] - 1;
if (sz[k] == 1)
{
ko+=dfs1(k);
}
}
u = min(u, ou + ans);
}
}
cout << u << endl;
}
}
