Hdu 1003 Max Sum
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 246702 Accepted Submission(s): 58259
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> int main() { int T; int k; scanf("%d", &T); for (k = 1; k <= T; k++) { int N; int a; int i; int start, end, pos; int temp, result; scanf("%d", &N); scanf("%d", &a); start = end = pos = 0; temp = result = a; for (i = 1; i < N; i++) { scanf("%d", &a); if (temp + a < a) { temp = a; pos = i; } else { temp += a; } if (temp > result) { result = temp; start = pos; end = i; } } printf("Case %d:\n", k); printf("%d %d %d\n", result, start+1, end+1); if (k != T) printf("\n"); } return 0; }