SDAU 课程练习3 1001

Problem A

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 56   Accepted Submission(s) : 8

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.<br>

 

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).<br>

 

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.<br>

 

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

 

题目大意：

求出最大连续子序列的和，并且给出起始位置和终止位置

思路：

前面博客已经写过了这个。不再写。

感想：

啦啦啦~刚才c++逃课被点名了。。。以后都不能逃c++了，好难过。。。

AC代码：

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
int j,i,k,n,m,t;
int a[100002];
scanf("%d",&t);
for (j=1;j<=t;j++)
{
scanf("%d",&n);
for (i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int sum=0,maxsum=-1001,first =0, last = 0, temp = 1;
for (i=0;i<n;i++)
{
sum += a[i];
if (sum > maxsum)
{
maxsum = sum;first = temp;last = i+1;
}
if (sum < 0)
{
sum = 0;temp = i+2;
}
}

printf("Case %d:\n%d %d %d\n",j,maxsum,first,last);
if (j!=t)
{
printf("\n");
}
}

return 0;
}