zoukankan      html  css  js  c++  java
  • codeforces 653C C. Bear and Up-Down(乱搞题)

    题目链接:

    C. Bear and Up-Down

     

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    The life goes up and down, just like nice sequences. Sequence t1, t2, ..., tn is called nice if the following two conditions are satisfied:

    • ti < ti + 1 for each odd i < n;
    • ti > ti + 1 for each even i < n.

    For example, sequences (2, 8), (1, 5, 1) and (2, 5, 1, 100, 99, 120) are nice, while (1, 1), (1, 2, 3) and (2, 5, 3, 2) are not.

    Bear Limak has a sequence of positive integers t1, t2, ..., tn. This sequence is not nice now and Limak wants to fix it by a single swap. He is going to choose two indices i < j and swap elements ti and tj in order to get a nice sequence. Count the number of ways to do so. Two ways are considered different if indices of elements chosen for a swap are different.

    Input

    The first line of the input contains one integer n (2 ≤ n ≤ 150 000) — the length of the sequence.

    The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 150 000) — the initial sequence. It's guaranteed that the given sequence is not nice.

    Output

    Print the number of ways to swap two elements exactly once in order to get a nice sequence.

    Examples
    input
    5
    2 8 4 7 7
    output
    2
    input
    4
    200 150 100 50
    output
    1
    input
    10
    3 2 1 4 1 4 1 4 1 4
    output
    8
    input
    9
    1 2 3 4 5 6 7 8 9
    output
    0
    Note

    In the first sample, there are two ways to get a nice sequence with one swap:

    1. Swap t2 = 8 with t4 = 7.
    2. Swap t1 = 2 with t5 = 7.

    In the second sample, there is only one way — Limak should swap t1 = 200 with t4 = 50.

    题意:问有多少种交换方法使序列满足nice的条件;

    思路:把不合理的位置找出来,然后暴力交换看有多少种方式,乱搞就能搞出来,然而我乱搞了一夜晚一直wa,后来好好想了不同的情况才过;

    AC代码:

    //代码写的跟翔一样,自己看了就想吐;

    #include <bits/stdc++.h> using namespace std; const int N=15e4+3; int a[N],flag[N],pos[N],cnt,num,n; int sap(int x,int y) { int t=a[y]; a[y]=a[x]; a[x]=t; } int check(int v) { /* if(x==3) { for(int j=0;j<x;j++) { cout<<pos[j]<<"&"<<endl; } } */ int u=v; // cout<<v<<"@"<<pos[v-1]<<endl; for(int j=0;j<u;j++) { //cout<<a[1]<<"*"<<a[4]<<endl; //cout<<pos[j]<<"#"<<j<<endl; if(pos[j]==1) { if(a[pos[j]+1]<=a[pos[j]])return 0; continue; } if(pos[j]==n) { if(n%2) { if(a[n-1]<=a[n])return 0; continue; } else { if(a[n-1]>=a[n])return 0; continue; } } if(pos[j]%2) { if(a[pos[j]-1]<=a[pos[j]]||a[pos[j]+1]<=a[pos[j]])return 0; } else { if(a[pos[j]-1]>=a[pos[j]]||a[pos[j]+1]>=a[pos[j]])return 0; } } return 1; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } a[0]=10000000; a[n+1]=0; cnt=0,num=0; for(int i=1;i<n;i++) { if(i%2) { if(a[i]>=a[i+1]){ if(!flag[i]){pos[num++]=i,flag[i]=1;} if(!flag[i+1])pos[num++]=i+1,flag[i+1]=1; } } else { if(a[i]<=a[i+1]) { if(!flag[i]) pos[num++]=i,flag[i]=1; if(!flag[i+1])pos[num++]=i+1,flag[i+1]=1; } } } if(num>=8){cout<<"0"<<endl;return 0;} int ans=0; for(int i=0;i<num;i++) { for(int j=i+1;j<num;j++) { sap(pos[i],pos[j]); ans+=check(num); sap(pos[j],pos[i]); } } for(int i=0;i<num;i++) { for(int j=1;j<=n;j++) { if(!flag[j]) { sap(pos[i],j); pos[num]=j; ans+=check(num+1); sap(j,pos[i]); } } } cout<<ans<<" "; return 0; }
  • 相关阅读:
    一本书
    在linux oracle 10g/11g x64bit环境中,goldengate随os启动而自己主动启动的脚本
    关于仿酷狗音乐播放器开源:寻求一套音乐播放器素材,让仿酷狗开源
    Cocos2d-x中Vector&lt;T&gt;容器以及实例介绍
    成都传智播客Java/PHP培训就业率高
    P1341 无序字母对
    P1168 中位数
    P1146 硬币翻转
    P1340 兽径管理
    P2023 [AHOI2009]维护序列
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5301401.html
Copyright © 2011-2022 走看看