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
和最大的最长子序列:#includeint main (){ int T, i, n, x, a, b, c, sum, M, k = 0, flag = 0; scanf("%d", &T); while (T--) { k++; scanf("%d %d", &n, &x); sum = M = x; a = b = c = 1; for (i = 2; i <= n; i++) { scanf("%d", &x); if (sum + x < x) { sum = x; a = i; } //若sum+x比x小,则将sum置为x,暂将此时的位置记录下来 else sum += x; //否则直接加上 if (sum > M) { M = sum; b = a; c = i; } //当找到更大子序列的和时更改最大值和对应位置 } if (flag == 1) printf("\n"); printf("Case %d:\n", k); printf("%d %d %d\n", M, b, c); flag = 1; } return 0;}