1007. Maximum Subsequence Sum (25)

Given a sequence of K integers { N₁, N₂, ..., N_K }. A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } 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 the last numbers of the whole sequence.

Sample Input:

10-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

问题分析：求最大子序列，方法分析比较请见：http://blog.csdn.net/biaobiaoqi/article/details/8762618（1）动态规划或者分治都可以考虑。（2）本解参考网上一例，从题目中提取一条原则“若某段和小于0，则其不可能是最大子序列的首部分。（3）根据这一原则，可在一轮循环内完成处理。

#include <stdio.h>#include <iostream>using namespace std;int main(int argc, char *argv[]){int K;int i;int tempSum=0,tempBegin=0,tempEnd=0;int sum=0,begin=0,end=0;cin>>K;int *ps=new int [K];for (i=0;i<K ;i++ ){cin>>ps[i];}for (i=0;i<K ;i++ ){if (tempSum>=0){tempSum=tempSum+ps[i];tempEnd=i;}else {tempSum=0;tempSum +=ps[i];tempBegin=i;tempEnd=i;}if (tempSum>sum||(tempSum==0&&end==K-1)){sum=tempSum;begin=tempBegin;end=tempEnd;}}    cout<<sum<<" "<<ps[begin]<<" "<<ps[end]<<endl;delete [] ps;return 0;}