HDU.1003【max-sum】---16.2.3

第一次开通博客，想以这样一个平台来记录我接下来的ACM的奋斗之路。

虽然是女生，但是我相信男生可以的我相信我就可以。

在不知道对手的情况下，只要奋发努力尽力做到最好就可以了。

不说废话了，切入主题：

Max Sum

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

Total Submission(s): 197539 Accepted Submission(s): 46151



Problem Description

Given a sequence a[1],a[2],a[3]......a
, 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

代码如下：

#include <stdio.h>

int a[100001];

int main()

{

int casenumber, number, k = 0, i;

int position, xbegin, xend;

int maxsum, thissum;

scanf("%d",&casenumber);

while(casenumber--)

{

k ++;

scanf("%d",&number);

for(i = 1; i <= number; i ++)

{

scanf("%d", &a[i]);

}

position = 1;

xbegin = xend = 0;

thissum = 0;

maxsum = -9999;

for(i = 1; i <= number; i++)

{

thissum += a[i];

if(thissum > maxsum)

{

maxsum=thissum;

xbegin=position;

xend=i;

}

if(thissum < 0)

{

thissum = 0;

position = i+1;

}

}

printf("Case %d:\n%d %d %d\n",k,maxsum,xbegin,xend);

if(casenumber)//这里去掉 -1，因为while先判断了，然后执行casenumber=casenumb-1，最后一次的时候casenumber已经等于0了

printf("\n");

}

return 0;

}

总结：这道题的基本思想是求最大子序列，用每个数加进去都计算一遍的方法：1.若当前和大于最大和（xbegin始终等于posion）：改变最大和的值,同时改变begin,end的值，当前位置position就是最大子序列的第一个数的位置，i移动到哪里，哪里就是当前最大子序列最后一个位置。2.若前i项为负的，从第i+1项重新开始做（position
= i+1，thissum = 0）：前面的和若为负，再加谁，就相当于a[i]减去了前面的数，和一定会减小，此时只需将position放到下一位。

还要多谢我哥们不厌其烦的给我改代码【感动脸】