sdut-1351-Max Sum-hdu-1003

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本文介绍了一种求解整数序列中最大和子序列的算法，并提供了完整的C语言实现代码。通过对输入序列的逐个元素扫描，算法能够高效地找到具有最大和的连续子序列及其起始和终止位置。

题目描述

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.

输入

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).

输出

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.

示例输入

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

示例输出

Case 1:
14 1 4

Case 2:
7 1 6

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int ls[100005];
int begin[100005];
int main()
{
    int t;
    scanf("%d",&t);
     for(int r = 1; r<=t; r++)
    {
        int n;
        scanf("%d",&n);
        int i;
        for(i = 1; i <= n; i++)
            scanf("%d",&ls[i]);
        int max= 0, sum = 0;
        int top = 0, end = 1;
        max = ls[1];
        sum = ls[1];
        begin[++top] = 1;
        for(i = 2; i <= n; i++)
        {
            sum += ls[i];
            if(sum >= max)
            {
                end = i;
                max = sum;
            }
            else if(sum < 0)
            {
                sum = 0;
                begin[++top] = i+1;
            }
        }
        printf("Case %d:\n",r);
        while(begin[top]>end)
        {
            top--;
        }
        if(r < t)
        {
            printf("%d %d %d\n",max,begin[top],end);
            printf("\n");
        }
        else
        {
            printf("%d %d %d\n",max,begin[top],end);
        }
    }
    return 0;
}