CF Round #569 Div2(contest1180)

Game link: http://codeforces.com/contest/1180

Problem A

Meaning of the questions: given n, the number of squares asked. Figure understood. . .

Solution:

Law can look for the found block number is 2n * (n-1) +1

Code:

#include<bits/stdc++.h>
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
    int n=read();
    printf("%d\n",2*n*n-2*n+1);
    return 0;
}

Problem B

Question is intended: to give you a sequence, where you can put it into any element opposite of -1, that: \ (a_i:--a_i. 1 \) , do the most program product

Solution：

Apparently as many non-negative number to perform this operation is optimal, and we are bound to make this the largest non-negative product

While for two non-negative \ (a, b (b> a) \) is, \ ((A +. 1) B> A (B +. 1) \)

So we only need to sequence the smallest negative value after the operation to perform and compare the size of the largest non-negative on the line

Code：

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+1;
int n,_minus,add,a[N];
int maxm,maxa=-1,idm,ida;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
    n=read();
    for(int i=1;i<=n;i++){
        a[i]=read();
        a[i]>=0?++add:++_minus;
        if(a[i]>=0&&a[i]>maxa) ida=i,maxa=a[i];
        if(a[i]<0&&a[i]<maxm) idm=i,maxm=a[i];
    }int f1=_minus%2,f2=add%2;
    if(f1==f2){
        for(int i=1;i<=n;i++)
            printf("%d ",a[i]<0?a[i]:-a[i]-1);
        return 0;
    }
    if(_minus==1&&!add){
        printf("%d ",-a[1]-1);
        return 0;
    }
    if(f1!=f2){
        int id,flag=0;
        if(-maxm-1>maxa) id=idm;
        else id=ida,flag=1;
        for(int i=1;i<id;i++)
            printf("%d ",a[i]<0?a[i]:-a[i]-1);
        printf("%d ",flag?a[id]:-a[id]-1);
        for(int i=id+1;i<=n;i++)
            printf("%d ",a[i]<0?a[i]:-a[i]-1);      
    }
}

Problem C

Question is intended: to give you a sequence, each operation sequence before removing the two elements \ (A, B \) , the larger elements in the first sequence, the smaller elements at the end of the sequence, there are m times query, each query you x-th operation out of which two elements

Solution:

Obviously, when the largest element in the sequence as the first, all the elements will be compared with it the tail kicked, form a cycle, and the replacement element to the head of the queue up to the maximum required operations n-1, then we can simulate the largest element replacement to all operations prior to the first team, after the operation, it can (1) is calculated according to the cycle O

Code:

#include<bits/stdc++.h>
#define ll long long 
#define mp make_pair
#define fir first
#define sec second
using namespace std;
const int N=1e5+1;
int n,m,maxn;
int head,cnt,tail,q[N*31];
pair<int,int> u[N];
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
    n=read(),m=read();
    for(int i=1;i<=n;i++){
        q[++tail]=read();
        if(q[i]>maxn) maxn=q[i];
    }head=1;
    while(1){
        int x=q[head],y=q[head+1];
        u[++cnt]=mp(x,y);
        if(x==maxn) break;
        if(x>y) q[head+1]=q[head++],q[++tail]=y;
        else ++head,q[++tail]=x;
    }
    for(int i=1;i<=m;i++){
        ll x=read();
        if(x<=cnt) printf("%d %d\n",u[x].fir,u[x].sec);
        else printf("%d %d\n",maxn,q[head+1+(x-cnt)%(n-1)]);
    }
    return 0;
}

Problem D

The meaning of problems: Given a matrix of n * m, you start a (1,1); every movement, set point is your location (x, y), you can select any tuple (dx, dy) , moved to (x + dx, y + dy) points, but each selected tuple (dx, dy) can not be used before. Now ask whether there is a plan to make you just go through each point in time, if there is, according to the order reaches the output point, or output -1.

Solution:

This is a configuration problem. From the (1,1) and (n, m) of the counter-shape round two jumping points sequentially, easy to prove correctness, Refer to code (described as not very good)

Code:

#include<bits/stdc++.h>
#define mp make_pair
using namespace std;
const int N=1e6+1;
int n,m,add1=1,add2=-1;
int nx,ny,mx,my,flag;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
void ins(int opt){
    if(!opt){
        printf("%d %d\n",nx,ny);
        if(nx+add1>n||nx+add1<1) ++ny,add1=-add1;
        else nx+=add1;
    }else{
        printf("%d %d\n",mx,my);
        if(mx+add2<1||mx+add2>n) --my,add2=-add2;
        else mx+=add2;
    }
}
int main(){
    n=read(),m=read();
    nx=ny=1,mx=n,my=m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            ins(flag),flag^=1;
    return 0;
}