今日头条2018校园招聘第一题 ---POJ 2479

[align=left]第一次参加公司的招聘笔试，虽然只是抱着试试水的心态去参加的，可惜的是第一题就做错了。。。。。[/align]
[align=left]第一题，其实只是一个求最大子段和的变式题，不过笔试的时候也不知道怎么了，就是不知道思路，最后还写了一个错的思路[/align]
[align=left]题目大意：笔试题是求两个不相邻区间的最大子段和， OJ题是不相交区间的最大子段和
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[align=left]思路：很简单，遍历一遍，记录从前往后每个元素以该元素结尾的最大子段和，从后往前每个元素以该元素结尾的最大字段和，[/align]
[align=left]然后一层遍历枚举每一个断点，求出该断点前的最大子段和和断点后的最大子段和最大值
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[align=center]Maximum sum[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41980		Accepted: 13098

Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:



Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.

Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.
Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41980		Accepted: 13098

Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:



Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.

Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<sstream>
#include<cctype>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int INF=0x3f3f3f3f;
const int maxn=1234;

int T;
int a[50005];
int b1[50005],b2[50005];
int main()
{
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);

int sum=0;
int maxs=a[0];
for(int i=0;i<n;i++)、、从前往后的最大子段和
{
sum+=a[i];
if(sum>maxs) maxs=sum;
if(sum<0) sum=0;

b1[i]=maxs;
}
sum=0;
maxs=a[n-1];
for(int i=n-1;i>=0;i--){//从后往前的最大子段和
sum+=a[i];
if(sum>maxs) maxs=sum;
if(sum<0) sum=0;
b2[i]=maxs;
}

int ans=b1[0]+b2[1];
for(int i=0;i<n-1;i++)
{
ans=max(ans,b1[i]+b2[i+1]);
}

printf("%d\n",ans);

}

return 0;
}