time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard 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.
Examples input 5 1 0 0 1 0 output 4 input 4 1 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.
可以对一段区间里的01取反一次，n<=100， 1.暴力枚举区间在计算答案O(n^3)
2.暴力枚举区间前缀和优化算答案O(n^2)
3.观察每次用前缀和计算[L,R]答案(ai记录1-i中，1的个数,bi记录1-i中，0的个数） ans=a[L-1]+a[n]-a[R]+a[R]-b[L-1]; 现在考虑以R为右端点的区间，a[n]-a[R]+b[R]这一部分始终是相同的，所以我们只要找到a[L-1]-b[l-1]这段区间的最大值即可，而这个最大值可以直接开一个变量维护 复杂度O(n)
O(n^2)代码
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<iostream> using namespace std; const int maxn=1e5+5; int a[maxn],b[maxn],n,x; int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;++i) { scanf("%d",&x); a[i]=a[i-1]+(x==1); b[i]=b[i-1]+(x==0); } int ans=0; for(int i=1;i<=n;++i) for(int j=i;j<=n;++j) ans=max(ans,a[i-1]+a[n]-a[j]+b[j]-b[i-1]); //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]); printf("%d\n",ans); } return 0; }O(n)代码
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<iostream> using namespace std; const int maxn=1e5+5; int a[maxn],b[maxn],n,x; int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;++i) { scanf("%d",&x); a[i]=a[i-1]+(x==1); b[i]=b[i-1]+(x==0); } int ans=0,res=0; for(int i=1;i<=n;++i) { ans=max(ans,res+a[n]-a[i]+b[i]); if(a[i]-b[i]>res)res=a[i]-b[i]; } //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]); printf("%d\n",ans); } return 0; }