Max sum。。。。。hdu oj----1003
Max sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 111859 Accepted Submission(s): 25831
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 int main() 3 { 4 int T,N,count=0,a[100010]; 5 scanf("%d",&T); 6 while(T--) 7 { 8 count++; 9 int start,s,end,e,max,sum,i; 10 scanf("%d",&N); 11 for(i=1;i<=N;i++) 12 scanf("%d",&a[i]); 13 /*此题主要意思是记录任意几个连续数字的和最大,并记录开始与结束的位置*/ 14 for(sum=max=a[1],s=e=start=end=1,i=2;i<=N;i++) 15 { /*开始与结束的位置是变化的*/ 16 if(sum<0){sum=0;start=i;end=i-1;} 17 /*第一种情况,sum变为负数时,sum归零处理不会影响后面的判断,起始值变为i,末尾值变为i-1*/ 18 sum+=a[i]; end++; 19 if(sum>max){s=start;e=end;max=sum;} 20 /*第二种情况,s 记录当前的起始值,e 记录当前的末位置*/ 21 } 22 printf("Case %d:\n",count); 23 printf("%d %d %d\n",max,s,e); 24 if(T>0) printf("\n"); 25 } 26 return 0; 27 }