HDU 1003 Max Sum (O（N）算法)

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本文解析了一个经典的编程问题——最大子序列和问题，通过示例详细介绍了如何找到数组中具有最大和的连续子序列及其起始和结束位置。文章提供了一种高效的遍历算法实现。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 210088 Accepted Submission(s): 49285

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

大体题意：
给你n 个元素的数组，求出一个连续子序列 使得子序列的和最大，并且输出 子序列的起始位置和终止位置！

思路：
直接遍历数组，thissum += a[i]，当比maxsum 大时，更新maxsum 并且更新左位置和右位置
如果thissum < 0肯定不能作为起点，也不是最优的直接把thissum初始化为0，然后更新左位置！
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn];
int l,r,n,k;
int solve(){
    l = r = k = 1;
    int maxsum = -maxn,thissum = 0;
    for (int i = 1; i <= n; ++i){
        thissum += a[i]; 
        if (thissum > maxsum)maxsum = thissum,r=i,l=k;
        if (thissum < 0)thissum = 0,k=i+1;
    }
    return maxsum;
}
int main(){
    int T,cnt=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for (int i = 1; i <= n; ++i){
            scanf("%d",&a[i]);
        }
        int ans = solve();
        if (cnt)printf("\n");
        printf("Case %d:\n",++cnt);
        printf("%d %d %d\n",ans,l,r);
    }
    return 0;
}