Codeforces Round #553 (Div. 2) Solution

This morning opened a \ (\ RM vp \) , the first four questions is water, looked \ (\ rm E, F \ ) after me to write a \ (\ RM F \) , and then not tune out of blood loss ... After the game only to find \ (\ rm E \) is really water ...

Game Portal: https://codeforces.com/contest/1151

A. Maxim and Biology

https://codeforces.com/contest/1151/problem/A

Direct violence just fine.

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf long double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-12;
const int mod = 1e9+7;

int n;
char s[maxn];

int get(int a,int b) {
    return min(abs(a-b),26-max(a,b)+min(a,b));
}

signed main() {
    read(n);scanf("%s",s+1);
    int ans=1e9;
    for(int i=1;i<=n-3;i++) ans=min(ans,get(s[i],'A')+get(s[i+1],'C')+get(s[i+2],'T')+get(s[i+3],'G'));
    write(ans);
    return 0;
}

B. Dima and a Bad XOR

https://codeforces.com/contest/1151/problem/B

This problem is very strange ... I just fucks ...

We consider each column random number out of a random ten thousand times, the probability of error tends to \ (0 \) a.

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf long double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-12;
const int mod = 998244353;
const int inv2 = 499122177;

int n,m,a[505][505],t[505];

signed main() {
    srand(time(0));
    read(n),read(m);FOR(i,1,n) FOR(j,1,m) read(a[i][j]);
    for(int i=1;i<=20000;i++) {
        int res=0;
        for(int j=1;j<=n;j++) res^=a[j][t[j]=rand()%m+1];
        if(res) {
            puts("TAK");
            for(int k=1;k<=n;k++) printf("%d ",t[k]);
            puts("");return 0;
        }
    }puts("NIE");
    return 0;
}

C. Problem for Nazar

https://codeforces.com/contest/1151/problem/C

Multiplication, and then press the meaning of the questions simulation.

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf long double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-12;
const int mod = 1e9+7;
const int inv2 = 5e8+4;

int a[2];

int sum(int x,int y,int t) {
    int qwe=(y-x+1)%mod;
    x<<=1,y<<=1;if(t) x--,y--;
    x%=mod,y%=mod;
    return 1ll*(x+y)*qwe%mod*inv2%mod;
}

int solve(int r) {
    int tot=0,now=1,kd=1;
    int ans=0;a[0]=a[1]=1;
    while(tot<r) {
        if(tot+now>r) ans+=sum(a[kd],a[kd]+r-now,kd);
        else ans+=sum(a[kd],a[kd]+now-1,kd);
        a[kd]+=now,tot+=now,now<<=1,kd^=1;
    }return (ans%mod+mod)%mod; 
}

signed main() {
    int l,r;read(l),read(r);
    write((solve(r)-solve(l-1)+mod)%mod);
    return 0;
}

D. Stas and the Queue at the Buffet

https://codeforces.com/contest/1151/problem/D

The formula is open:
\ [nb_i-a_i + J \ CDOT (a_i-B_i) \]
Therefore, according to the direct greedy \ (a_i-b_i \) sorting like.

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf long double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-12;
const int mod = 998244353;
const int inv2 = 499122177;

int n,a[maxn],b[maxn],t[maxn];

signed main() {
    read(n);FOR(i,1,n) read(a[i]),read(b[i]),t[i]=a[i]-b[i];
    sort(t+1,t+n+1);int res=0;
    for(int i=n;i;i--) res+=t[n-i+1]*i;
    for(int i=1;i<=n;i++) res+=-1*a[i]+n*b[i];
    write(res);
    return 0;
}

E. Number of Components

https://codeforces.com/contest/1151/problem/E

Noting forest communication block number \ (= \) Points \ (- \) number of sides.

Enumeration of each point and each edge contributions on the line.

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
#define ll long long

void print(ll x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(ll x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-8;

ll ans,s;
int n,a[maxn];

int c(int x) {return x*(x-1)/2+x;}

signed main() {
    read(n);FOR(i,1,n) read(a[i]),ans+=c(n)-c(n-a[i])-c(a[i]-1);
    FOR(i,1,n-1) {
        int l=min(a[i],a[i+1]),r=max(a[i],a[i+1]);
        ans-=1ll*l*(n-r+1);
    }write(ans);
    return 0;
}

F. Sonya and Informatics

https://codeforces.com/contest/1151/problem/F

This is a bit difficult in this unique title of ...~~Perhaps I was too dishes right~~

The final state must be noted that the foregoing \ (K \) a \ (0 \) , followed by all \ (1 \) , where \ (K \) is a start \ (0 \) number.

We consider a violent \ (dp \) , set \ (f [i] [j ] \) representation of a \ (i \) operations, before \ (k \) positions have \ (j \) a \ (0 \) of the total number of cases.

And then transferred to enumerate several situations:

Front (front means \ (K \) a white th position) \ ((0) \) a black and behind \ ((. 1) \) \ (\ RM the swap \) , then the number in front of the white \ ( -1 \) , the program number \ (I (NK- (Ki)) \) .
Front and rear black white change, the number of white \ (1 + \) , the program number and the same reason as above.
The same color change, the same number of white.
And in front of or behind the front and rear of the transducer change, the same number of white.

Such complexity is transferred \ (O (NK) \) , but \ (k \) too had not.

Noting \ (n-\) only \ (100 \) we can optimize the transfer matrix above, the complexity of the \ (O (n-^. 3 \ log K) \) .

#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf long double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-12;
const int mod = 1e9+7;

int add(int x,int y) {return x+y>=mod?x+y-mod:x+y;}
int del(int x,int y) {return x-y<0?x-y+mod:x-y;}
int mul(int x,int y) {return 1ll*x*y-1ll*x*y/mod*mod;}
void inc(int &x,int y) {x+=y;x%=mod;}

int n,k,a[maxn],res,c;

int qpow(int a,int x) {
    int res=1;
    for(;x;x>>=1,a=1ll*a*a%mod) if(x&1) res=1ll*res*a%mod;
    return res;
}

struct Matrix {
    int a,b,r[110][110];

    Matrix () {a=b=0,memset(r,0,sizeof r);}
    
    Matrix operator * (const Matrix &t) const {
        Matrix res;res.a=a,res.b=t.b;
        for(int i=0;i<=a;i++)
            for(int j=0;j<=t.b;j++)
                for(int k=0;k<=b;k++)
                    res.r[i][j]=add(res.r[i][j],mul(r[i][k],t.r[k][j]));
        return res;
    }
}tr,ans;

signed main() {
    read(n),read(k);
    for(int i=1;i<=n;i++) read(a[i]),res+=!(a[i]&1);
    for(int i=1;i<=res;i++) c+=!a[i];
    ans.a=0,ans.b=res;ans.r[0][c]=1;tr.a=tr.b=res;
    for(int i=0;i<=res;i++) {
        int lw=i,lb=res-i,rw=res-lw,rb=n-res-lb;
        if(i!=res) inc(tr.r[i][i+1],lb*rw%mod);
        if(i!=0) inc(tr.r[i][i-1],lw*rb%mod);
        inc(tr.r[i][i],(res*(res-1)+(n-res)*(n-res-1))%mod*qpow(2,mod-2)%mod);
        inc(tr.r[i][i],(lw*lb+rw*rb)%mod);
    }
    int x=k,f=0;
    for(;x;x>>=1,tr=tr*tr) if(x&1) ans=ans*tr;
    FOR(i,0,res) f=(f+ans.r[0][i])%mod;write(ans.r[0][res]*qpow(f,mod-2)%mod);
    return 0;
}