Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 134690 Accepted Submission(s): 31181
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
Author
Ignatius.L
Recommend
简单dp思想：
1 //15 MS 236 KB Visual C++ 2 #include<stdio.h> 3 int main(void) 4 { 5 int t,n,k=1; 6 int a; 7 scanf("%d",&t); 8 while(t--) 9 { 10 if(k>1) printf(" "); 11 scanf("%d",&n); 12 int l=1,r=1,sum=-0x7ffffff,tmp=0,tl=1; 13 for(int i=1;i<=n;i++){ 14 scanf("%d",&a); 15 tmp+=a; 16 if(tmp>sum){ 17 sum=tmp; 18 l=tl; 19 r=i; 20 } 21 if(tmp<0){ 22 tmp=0; 23 tl=i+1; 24 } 25 } 26 printf("Case %d: ",k++); 27 printf("%d %d %d ",sum,l,r); 28 } 29 return 0; 30 } 31 /* 32 33 30 34 3 -1 0 -1 35 3 -1 0 1 36 3 -1 -2 -3 37 38 */