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
input
41 0 0 1
output
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，现允许你对任意一个子序列取反，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，保存一个最大值即可。
ac代码：
#include #include using namespace std;int a[105];int main(){// freopen(in.txt,r,stdin); int n; while(cin>>n){ for(int i =0; i>a[i]; int ans = 0; for(int i=0; i=i && k