Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 107407 Accepted Submission(s): 24742
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
1 #include <stdio.h> 2 3 int main() 4 { 5 int T, N; 6 scanf("%d", &T); 7 for(int i = 1; i <= T; i++) 8 { 9 int sum = 0, k = 1, nTemp; 10 int max_sum = -0x7ffffff; 11 int sSeq, eSeq; 12 13 scanf("%d", &N); 14 for(int j = 1; j <= N; j++) 15 { 16 scanf("%d", &nTemp); 17 sum += nTemp; 18 if(sum > max_sum) 19 { 20 max_sum = sum; 21 sSeq = k; 22 eSeq = j; 23 } 24 if(sum < 0) //如果总和小于0，则其实位置从下一个开始 25 { 26 k = j + 1; 27 sum = 0; 28 } 29 } 30 if(i != 1) 31 printf("\n"); 32 printf("Case %d:\n", i); 33 printf("%d %d %d\n", max_sum, sSeq, eSeq); 34 } 35 return 0; 36 }
总结：发现自己题目做的太少了，以至于连很简单的题目都做不出来，只能说自己还不够努力吧！