今日头条2018校园招聘第一题 ---POJ 2479

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。。。。。

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

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

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，

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

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal 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

1101 -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: 65536KTotal 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

1101 -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;}