Codeforces Round 979 (Div. 2)题解记录
比赛链接：https://codeforces.com/contest/2030
A. A Gift From Orangutan
肯定最小值和最大值放前面最好，答案得解
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(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 = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[150000];
int main()
{
fio();
ll t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
for(ll i=1;i<=n;i++)cin>>a[i];
sort(a+1,a+1+n);
ll sum=0;
for(ll i=2;i<=n;i++)
{
sum+=a[n]-a[1];
}
cout<<sum<<endl;
}
}
B. Minimise Oneness
浅浅枚举下，发现答案不可能为0。1随便放
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(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 = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[150000];
int main()
{
fio();
ll t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
for(ll i=1;i<=n/2;i++)
{
cout<<0;
}
cout<<1;
for(ll i=n/2+2;i<=n;i++)
{
cout<<0;
}
cout<<endl;
}
}
C. A TRUE Battle
and是优先于or的，考虑1如果在最左或者最右，一个or即可胜利，否则看是否有两个连续的1，有即胜利
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(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 = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[150000];
int main()
{
fio();
ll t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
string f;
cin>>f;
ll pd=0;
if(f[0]=='1'||f[f.size()-1]=='1')pd=1;
for(ll i=0;i<f.size()-1;i++)
{
if(f[i]=='1'&&f[i+1]=='1')pd=1;
}
if(pd)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
}
D. QED's Favorite Permutation
本题多个连续查询，于是得考虑连续维护；发现其实可以把数组分成多个不同得移动区间
，如果这个区间全是R或者L或者RRL这种单调的串就一定可以排好序
赛时只想到了线段树。
还有种方法,用空数组，对于L的位置++，R的位置就后面一个位置++，这样子去维护串，如果对于此区间（除了第一位）有0则始终都不可以排序
于是计不合规的0的个数即可
好友提供的非线段树版：
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<iomanip>
#include<numeric>
#include<stack>
#include<deque>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define endl "\n"
using namespace std;
ll t, n, a[200005], q, en[200005], now[200005];
string s;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> t;
while (t--)
{
cin >> n >> q;
fill(en + 1, en + n + 1, 0);
fill(now + 1, now + n + 1, 0);
for (int i = 1; i <= n; i++)
cin >> a[i];
cin >> s;
ll maxn = -1, be = -1;
for (int i = 1; i <= n; i++)
{
if (a[i] != i)
{
if (be == -1)
be = i;
maxn = max(maxn, a[i]);
}
if (i == maxn)
{
for (int j = be; j < i; j++)
{
en[j] = 1;
}
be = -1;
maxn = -1;
}
}
for (int i = 0; i < n; i++)
{
if (s[i] == 'L')
now[i]++;
else
now[i + 1]++;
}
ll ch = 0;
for (int i = 1; i < n; i++)
{
if (en[i] && !now[i])
ch++;
}
while (q--)
{
ll p;
cin >> p;
p--;
if (s[p] == 'L')
{
s[p] = 'R';
now[p]--;
now[p + 1]++;
if (en[p] && !now[p])
ch++;
if (en[p + 1] && now[p + 1] == 1)
ch--;
}
else
{
s[p] = 'L';
now[p + 1]--;
now[p]++;
if (en[p + 1] && !now[p + 1])
ch++;
if (en[p] && now[p] == 1)
ch--;
}
cout << (ch ? "NO" : "YES") << endl;
}
}
return 0;
}
线段树版：
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
// const ll p=rnd()%mod;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(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 = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
const ll maxn = 2e5+5;
struct s
{
ll l, r;
ll ad, v, cf;//ad为加法的懒惰标记，cf为乘法的懒惰标记
ll la;
}p[maxn << 2];
ll fa[maxn << 2];
void build(ll i, ll l, ll r)
{
p[i].l = l,p[i].r = r,p[i].v = 0;
p[i].la=-1;
if (l == r)
{
fa[l] = i;
return;
}
build(i << 1, l, (l + r) >> 1);
build(i << 1 | 1, ((l + r) >> 1) + 1, r);
}
void push_down(ll i)
{
if (p[i].la!=-1)
{
p[i].v=(p[i].r-p[i].l+1)*p[i].la;
p[i<<1].v=(p[i<<1].r-p[i<<1].l+1)*p[i].la;
p[i<<1|1].v=(p[i<<1|1].r-p[i<<1|1].l+1)*p[i].la;
p[i<<1].la=p[i].la;
p[i<<1|1].la=p[i].la;
p[i].la=-1;
}
}
void udq(ll i, ll ad, ll cf, ll l, ll r)
{
if (p[i].l == l && p[i].r == r)
{
p[i].v=ad*(r-l+1);
p[i].la=ad;
return;
}
push_down(i);
ll i1 = i << 1, i2 = i << 1 | 1;
if (l <= p[i1].r)
{
if (r <= p[i1].r)
udq(i1, ad, cf, l, r);
else
udq(i1, ad, cf, l, p[i1].r);
}
if (r >= p[i2].l)
{
if (l >= p[i2].l)
{
udq(i2, ad, cf, l, r);
}
else
udq(i2, ad, cf, p[i2].l, r);
}
p[i].v = p[i1].v + p[i2].v;
}
ll query(ll i, ll l, ll r)
{
ll ans = 0;
if (p[i].l == l && p[i].r == r)
{
ans = p[i].v;
return ans;
}
push_down(i);
ll i1 = i << 1, i2 = i << 1 | 1;
if (l <= p[i1].r)
{
if (r <= p[i1].r)
ans = ans + query(i1, l, r);
else
ans = ans + query(i1, l, p[i1].r);
}
if (r >= p[i2].l)
{
if (l >= p[i2].l)
{
ans = ans + query(i2, l, r);
}
else
ans = ans + query(i2, p[i2].l, r);
}
return ans;
}
ll a[1800000];
ll b[1800000];
bool vis[1800000];
ll d[2500000];
map<ll,pair<ll,ll>>q;
set<ll>pk;
int main()//发现RRR or LLL or RRRLL都是对的，每次只要对应处理区间问题即可
{
fio();
ll t;
cin>>t;
while(t--)
{
//cout<<66<<endl;
q.clear();
pk.clear();
ll n,m;
cin>>n>>m;
build(1,1,n);
//cout<<66<<endl;
for(ll i=1;i<=n;i++)cin>>a[i],b[i]=i,vis[i]=0,d[i]=0;
string f;
map<ll,ll>ok;
cin>>f;
ll cnt=0;
ll l=0;
ll op=0;
for(ll i=1;i<=n;i++)
{
if(cnt==0&&a[i]==i)continue;
if(a[i]!=i)
{
cnt=max(cnt,a[i]);
if(l==0)
{
l=i;
}
}
if(cnt==i)
{
cnt=0;
op++;
q[op]={l,i};//区间
ll o=0;
for(ll j=l;j<=i;j++)
{
b[j]=op;
vis[j]=1;
if(f[j-1]=='R')o++;
}
ok[op]=o;
//pk.insert();
l=0;
}
}
//cout<<66<<endl;
for(ll i=0;i<f.size();i++)
{
if(f[i]=='L')udq(1,0,0,i+1,i+1);
else udq(1,1,0,i+1,i+1);
}
//cout<<66<<endl;
cnt=0;
//cout<<op<<endl;
for(ll i=1;i<=op;i++)
{
ll c=ok[i];
ll l=q[i].first;
ll r=q[i].second;
//cout<<c<<" "<<r<<" "<<l<<endl;
if(c==r-l+1)continue;
ll j=(r-l+1)-c;
if(query(1,r-j+1,r)==0)
{
continue;
}
else cnt++;
}
//cout<<cnt<<endl;
while(m--)
{
ll wz;
cin>>wz;
if(vis[wz]==0)
{
if(cnt==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
else
{
if(f[wz-1]=='L')
{
f[wz-1]='R';
ll c=ok[b[wz]];
ll l=q[b[wz]].first;
ll r=q[b[wz]].second;
ll j=(r-l+1)-c;
ll hl1=0;
if(c==r-l+1||query(1,r-j+1,r)==0)
{
hl1=1;
}
udq(1,1,0,wz,wz);
ok[b[wz]]++;
c++;
ll hl2=0;
j=(r-l+1)-c;
if(c==r-l+1||query(1,r-j+1,r)==0)
{
hl2=1;
}
if(hl1==0&&hl2==1)cnt--;
else if(hl1==1&&hl2==0)cnt++;
if(cnt==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
else
{
f[wz-1]='L';
ll c=ok[b[wz]];
ll l=q[b[wz]].first;
ll r=q[b[wz]].second;
ll j=(r-l+1)-c;
ll hl1=0;
if(c==r-l+1||query(1,r-j+1,r)==0)
{
hl1=1;
}
udq(1,0,0,wz,wz);
ok[b[wz]]--;
c--;
ll hl2=0;
j=(r-l+1)-c;
if(c==r-l+1||query(1,r-j+1,r)==0)
{
hl2=1;
}
if(hl1==0&&hl2==1)cnt--;
else if(hl1==1&&hl2==0)cnt++;
if(cnt==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
}
}
}
}
