Flipping Game（Codeforces）(枚举)

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

51 0 0 1 0

output

4

input

41 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.

就是一个枚举题目：求怎样翻一个区间使得1的数量最多，输出1最多时的个数；

#include<stdio.h>#include<string.h>#include<stdlib.h>int a[151],b[151];int n,m;int main(){    while(scanf("%d",&n)!=EOF)    {        int max = 0;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(a[i] == 1)            {                max++;            }        }        if(max == n)        {            printf("%d\n",max-1);            continue;        }        max = 0;        for(int i=1;i<=n;i++)        {            for(int j=0;j<n;j++)            {                int count = 0;                for(int k=0;k<n;k++)                {                    b[k] = a[k];                }                for(int k=j;k<j+i;k++)                {                    if(j+i>n)                    {                        break;                    }                    if(b[k] == 1)                    {                        b[k] = 0;                    }                    else                    {                        b[k] = 1;                    }                }                for(int k=0;k<n;k++)                {                    if(b[k] == 1)                    {                        count++;                    }                }                if(max < count)                {                    max = count;                }            }        }        printf("%d\n",max);    }    return 0;}