Flipping Game（枚举）

大家好，又见面了，我是全栈君。

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 x means 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

复制

Output

4

复制

Input

4
1 0 0 1

复制

Output

4

复制

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.

题意：有n张牌，仅仅有0和1，问在[i，j]范围内翻转一次使1的数量最多。

输出1最多的牌的数量

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
    int n,i,j,k,t;
    int a[110];
    int sum[2];
    int cnt=0;
    while(~scanf("%d",&n))
    {
        cnt=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==1)
                cnt++;//记录開始时1的牌数
        }
        t=cnt;
        if(cnt==n)
        {
            printf("%d\n",n-1);//假设全是1的话 你得翻一张牌 所以剩下的最大数为总数-1
        }
        else
        {

            for(i=0; i<n; i++)
                for(j=i; j<n; j++)
                {
                    memset(sum,0,sizeof(sum));
                    for(k=i; k<=j; k++)
                        sum[a[k]]++;
                    if(sum[0]>sum[1])
                        {
                            if(cnt<t+sum[0]-sum[1])
                            {
                                cnt=t+sum[0]-sum[1];
                            }
                        }
                }
            printf("%d\n",cnt);
        }
    }
    return 0;
}

复制

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116604.html原文链接：https://javaforall.cn