枚举
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Description
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 valuesak for which their positions are in range[i, j] (that is i ≤ k ≤ j).
Flip the value of x means to apply operationx = 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 aren 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 Input
51 0 0 1 0
4
41 0 0 1
4
Sample Output
Hint
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.
分析:n值不大,暴力破解就行,用三次循环,第一层从0~n移动 i,第二层移动 j, 第三层移动范围。
#include<iostream>#include<string.h>#include<stdio.h>#include<algorithm>using namespace std;int main(){ int a[105],n,num[2]; cin>>n; int ans=0; for(int i=0;i<n;i++) { cin>>a[i]; if(a[i]==1)ans++; } int pos=ans; if(pos==n) printf("%d\n",n-1); else { for(int i=0;i<n;i++) for(int j=i;j<n;j++) { memset(num,0,sizeof(num)); for(int k=i;k<=j;k++)num[a[k]]++;if(num[0]>num[1])ans=max(ans,pos+num[0]-num[1]); } printf("%d\n",ans); }}