1007. Maximum Subsequence Sum (25)

Analysis:

(1) calculate the maximum continuous subsequence, and finally output the maximum subsequence value and the starting and ending elements of the subsequence. If all arrays are negative, 0 and the first and last elements of the array are output.

(2) A special case. For example, the input is:

4

-1 0 0-1

The output is: 0 0 0

Description:

 

Given a sequence of K integers {N1, N2 ,..., NK }. A continuous subsequence is defined to be {Ni, Ni + 1 ,..., nj} where 1 <= I <= j <= K. the Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. for example, given sequence {-2, 11,-4, 13,-5,-2}, its maximum subsequence is {11,-4, 13} with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. the numbers must be separated by one space, but there must be no extra space at the end of a line. in case that the maximum subsequence is not unique, output the one with the smallest indices I and j (as shown by the sample case ). if all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and last numbers of the whole sequence.

Sample Input:
10
-10 1 2 3 4-5-23 3 7-21
Sample Output:
10 1 4

 

Reference code:

 

# Include <iostream> using namespace std; int main () {int N; int * input; int I; int begin = 0, end = 0, sum = 0; // start position, end position, and subsequence sum of the final child sequence. Int tempSum = 0, tempBegin = 0, tempEnd = 0; // the starting position, ending position, and sum of subsequences currently being investigated. Cin> N; input = new int [N]; for (I = 0; I <N; I ++) cin> input [I]; end = N-1; for (I = 0; I <N; I ++) {if (tempSum> = 0) {tempSum + = input [I]; tempEnd = I ;} else {// If tempSum <0, then tempSum + input [I] <input [I] // at this point, we will start to evaluate the new subsequence tempSum = 0; tempSum + = input [I]; tempBegin = I; tempEnd = I;} // if (tempSum> sum) cannot be written as follows: if (tempSum> sum | (tempSum = 0 & end = N-1) {sum = tempSum; begin = tempBegin; end = tempEnd ;}} cout <sum <"" <input [begin] <"" <input [end] <endl; return 0 ;}