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  • Codeforces Round #191 (Div. 2)---A. Flipping Game

    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Iahub got bored, so he invented a game to be played on paper.

    He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of xmeans to apply operation x = 1 - x.

    The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

    Input

    The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.

    Output

    Print an integer — the maximal number of 1s that can be obtained after exactly one move.

    Sample test(s)
    input
    5
    1 0 0 1 0
    
    output
    4
    
    input
    4
    1 0 0 1
    
    output
    4
    
    Note

    In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

    In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.







    解题思路:题意是讲有一个0,1序列,现同意你对随意一个子序列取反,问操作后得到的序列,最多能有多少个1。

    数据不大,直接暴力。直接两层循环枚举取反序列的起点和终点,统计每种情况下1的个数,保存一个最大值就可以。






    AC代码:

    #include <iostream>
    #include <cstdio>
    using namespace std;
    
    int a[105];
    
    int main(){
    //  freopen("in.txt","r",stdin);
        int n;
        while(cin>>n){
            for(int i =0; i<n; i++)
                cin>>a[i];
            int ans = 0;
            for(int i=0; i<n; i++){
                for(int j=i; j<n; j++){
                    int sum = 0;
                    for(int k=0; k<n; k++){
                        if(k>=i && k<=j) sum += a[k]^1;          //用异或^取反
                        else sum += a[k];
                    }
                    if(ans < sum)  ans = sum;
                }
            }
            cout<<ans<<endl;
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4192380.html
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