Codeforces Round #569 (Div. 2)

http://codeforces.com/contest/1180/problems

B. Nick and Array

题意:给出一个n位序列存在一个变换使得 an变为 -an-1 求经过若干变换后序列的乘积最大

这道水题卡了半天实在不应该主要卡在 0 和-1 分了很多类

其实可以先默认全部变为负数如果此时个数为偶数那么就是最大值

如果为奇数：将最小的负数变为正数即可

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=1e5+10;
int n,m;
struct node
{
    int v,pos;
}s[N];
bool cmp(node a,node b)
{
    return a.v<b.v;
}
bool cmp2(node a,node b)
{
    return a.pos<b.pos;
}
int main()
{
    RI(n);
    rep(i,1,n)
    {
        int x;RI(x);if(x>=0)x=-x-1;
        s[i].v=x;s[i].pos=i;
    }
    if(n%2==0)
    {
        rep(i,1,n)
        printf("%d ",s[i].v);
    }
    else
    {
        sort(s+1,s+1+n,cmp);
        s[1].v=-s[1].v-1;
        sort(s+1,s+1+n,cmp2);
        rep(i,1,n)
        printf("%d ",s[i].v);
    }
    return 0;
}

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C. Valeriy and Deque

题意：给出n个数的序列 m个询问

将该序列看作双端队列每次从头取出两个数 A B 大的在前小的在后

问第 k 次取出的是哪两个数

一开始还模拟了一下找循环节。。。浪费了一些时间

其实当最大的在头就已经结束了循环节为n-1 然后模拟一下就可以了

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=3e5+10;
int n,m,x,maxx,ans[N][2],A,B,cnt;
deque<int>q;
ll qe[N];
int main()
{
    RII(n,m);
    int T=n-1;
    rep(i,1,n)
    {
        RI(x);
        q.push_back(x);
        maxx=max(maxx,x);
    }
    cnt=0;
    while(1)
    {
        ++cnt;
         A=q.front();q.pop_front();
         B=q.front();q.pop_front();
        ans[cnt][0]=A,ans[cnt][1]=B;
        if(A>B)
        q.push_front(A),q.push_back(B);
        else q.push_back(A),q.push_front(B);
        if(q.front()==maxx){break;}
    }
    int cnt2=cnt;
    rep(i,1,T)
    {
        ++cnt2;
        A=q.front();q.pop_front();
        B=q.front();q.pop_front();
        ans[cnt2][0]=A,ans[cnt2][1]=B;
        if(A>B)
        q.push_front(A),q.push_back(B);
        else q.push_back(A),q.push_front(B);
    }
    rep(i,1,m)
    {
        scanf("%lld",&qe[i]);
        if(qe[i]<=cnt2)printf("%d %d\n",ans[qe[i]][0],ans[qe[i]][1]);
        else
        {
            qe[i]=(qe[i]-cnt)%T;
            if(!qe[i])qe[i]=T;
            printf("%d %d\n",ans[ qe[i]+cnt ][0],ans[qe[i]+cnt][1]);
        }
    }
    return 0;
}

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D. Tolik and His Uncle

题意：给出一个n*m的矩阵一开始在1 1 求一种方法遍历所有的格子要求任意一步的行走向量不能相同如果走不完输出-1

显然一边从1 1 一边从n m 交替走蛇形就是答案

注意格子的奇偶情况

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=3e5+10;
int n,m;
int main()
{
    RII(n,m);
    int x=1,y=1,d1=1,x2=n,y2=m,d2=-1;
    if(n==1&&m==1){printf("1 1");return 0;}
    int X=n+1,Y=m+1;
    ll cnt=n*m;
    ll temp=cnt/2;

    while( temp--)
    {
        printf("%d %d\n",x,y);
        printf("%d %d\n",X-x,Y-y);
        y+=d1;
        if(y==m+1)
        {
            y=m;d1=-d1;x++;
        }
        if(y==0)
        {
            y=1;d1=-d1;x++;
        }
    }
    if(cnt%2==1)
    printf("%d %d",x,y);

    return 0;
}

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这题的输出量达到1000^2 经过测试 c 比c++ 快（当然 c++关闭了流）

<<endl <<'\n' 前者超时后者不超时

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=3e5+10;
int n,m;
int main()
{
    RII(n,m);
    int x=1,y=1,d1=1,x2=n,y2=m,d2=-1;
    if(n==1&&m==1){printf("1 1");return 0;}
    int X=n+1,Y=m+1;
    ll cnt=n*m;
    ll temp=cnt/2;

    while( temp--)
    {
        printf("%d %d\n",x,y);
        printf("%d %d\n",X-x,Y-y);
        y+=d1;
        if(y==m+1)
        {
            y=m;d1=-d1;x++;
        }
        if(y==0)
        {
            y=1;d1=-d1;x++;
        }
    }
    if(cnt%2==1)
    printf("%d %d",x,y);

    return 0;
}

View Code

最终rank 270 虽然打的很臭但居然是历史最好成绩了QAQ

Codeforces Round #569 (Div. 2)

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