Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 292444    Accepted Submission(s): 69379

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

Sample Output

Case 1: 14 1 4

Case 2: 7 1 6

题意

给出一个数组，求连续子段和的最大值

思路

第一次写用的前缀和，写完后发现复杂度太高了，交上去果断TLE

翻了一下以前写的51Nod的一道连续字段和的题，写了出来。复杂度O(n)

首先定义一个变量sum，使sum的值足够小，然后开始输入元素x，如果sum<0（即sum+x<x），将sum的值变为x，起点变为当前x的位置（相当于舍弃了前面的sum的值，从x开始重新累加）。然后定义ans，first，end分别记录最终的最大值，起点，终点。sum每改变一次，让ans和sum进行比较，如果ans<sum，更新ans的值，并将起点first为l，终点end变为当前x的位置i，最后输出即可。

一篇关于最大连续子序列和的讲的很不错的博客：https://blog.csdn.net/samjustin1/article/details/52043369。里面还有好多种写法，很详细

AC代码

#include<bits/stdc++.h>

#define ll long long

#define ms(a) memset(a,0,sizeof(a))

using namespace std;

const int maxn=1e6+10;

int main()

{

    ios::sync_with_stdio(false);

    int t;

    cin>>t;

    int _=0;

    while(t--)

    {

        int n;

        cin>>n;

        int m;

        ll ans=INT_MIN;

        ll sum=INT_MIN;

        int l;

        // sum，sum为临时记录最大值和起点位置的变量

        int first,end;

        for(int i=1;i<=n;i++)

        {

            cin>>m;

            //如果sum小于0，让sum从m重新开始

            //并改变l的值

            if(sum+m<m)

            {

                sum=m;

                l=i;

            }

            else

                sum+=m;

            // 如果ans小于sum，更新ans，first，end

            if(ans<sum)

            {

                first=l;

                end=i;

                ans=sum;

            }

        }

        cout<<"Case "<<++_<<':'<<endl;

        cout<<ans<<" "<<first<<" "<<end<<endl;

        // 注意输出。最后一组样例中没有空行，前面每组之间都有空行

        if(t)

            cout<<endl;

    }

    return 0;

}