cf div2 191 A
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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.
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.
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
51 0 0 1 0
4
41 0 0 1
4
一道水题,下面是代码:
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int main(){ int n; int a[10000]; while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++){ scanf("%d",&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){ if(a[k]==1){ sum++; } } else if(a[k]==0){ sum++; } } ans=sum>ans?sum:ans; } cout<<ans<<endl; } return 0;}
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