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Flipping Game
1 second
256 megabytes
standard input
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.
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.
5 1 0 0 1 0
4
4 1 0 0 1
4
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; }
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