Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 126443 Accepted Submission(s): 29293
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
#include<stdio.h>
void main()
{
int cases,i,m,n,sum,max,start,end,laststart,temp;
while(scanf("%d",&cases)!=EOF)
{
for(m=1;m<=cases;m++)
{
scanf("%d",&n);
sum=0;
max=-1001;//*****
start=end=laststart=0;
for(i=0;i<n;i++)
{
scanf("%d",&temp);
sum+=temp;
if(sum>max)
{
start=laststart;
end=i;
max=sum;
}
if(sum<0)
{
laststart=i+1;
sum=0;
}
}
printf("Case %d:\n",m);
printf("%d %d %d\n",max,start+1,end+1);
if(m!=cases)
printf("\n");
}
}
}
个人小结 1、设置一个laststart用来记录上一个开始的位置 2、sum设为0,max设为-1001,题目有数据范围-1000到1000