HDU 1003

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128040 Accepted Submission(s): 29659

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

Author

Ignatius.L

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题目说明：

看到题目的时候，把题目理解错了，一直看不懂题目是什么意思。

后来反复的读题目，原来是这个意思。（就是求和最大子序列）

拿题目中的输入输出例子举例吧：

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

其中，第一行的2：表示下面连续输入两行。

第二行的第一个数 5 ：表示这一行后面有5个数字要输入。（我就是这里理解错了）。

第三行的第一个数 7 表示后面有7个数字要输入。（实际数字为：0 6 -1 1 -6 7 -5 ，不包括前面的数字7）

该题我开始用动态规划+数组做的，估计是空间超出限制了，提交了7-8次，一直WA。

后来 无幻的博客 将数组去掉了，AC成功。

这里直接放出链接：http://blog.csdn.net/akof1314/article/details/4757021

/*Name: HDU 1003Copyright: Analyst Author: Analyst Date: 04/03/14 19:20Description: dev-cpp 5.5.3*/#include <stdio.h>int j, t;int max(int a, int b){return a >= b ? a : (t = j,b);      /*临时保存pos1*/}int main(){int T, N, i, maxValue;int temp, sum, pos1, pos2;scanf("%d", &T);for (i = 0; i < T; ++i){scanf("%d%d", &N, &temp);pos1 = pos2 = t = 1;maxValue = sum = temp;   for (j = 2; j <= N; ++j){scanf("%d",&temp);sum = max(sum+temp,temp);  /*如果相加之和小于该数*/if (sum >= maxValue)       {maxValue = sum;if (temp != 0)         /*避免情况：5 0 0 1 0 0*/pos1 = t;pos2 = j;}}printf("Case %d:\n%d %d %d\n",i+1,maxValue,pos1,pos2);if (i != T-1)printf("\n");}return 0;}

102198922014-03-04 23:20:51Accepted100315MS256K729 BC++Analyst

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