【PAT】1007. Maximum Subsequence Sum (25)
题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1007
分析：
（1）求最大连续子序列，最终输出最大子序列值以及子序列起始和终止元素。如果数组全为负数，则输出0以及数组的第一个和最后一个元素。
（2）一个特殊情况。如，输入的是：
4
-1 0 0 -1
那么我们输出的应该是： 0 0 0
题目描述：
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 the last numbers of the whole sequence.
Sample Input:10 -10 1 2 3 4 -5 -23 3 7 -21Sample Output:
10 1 4
参考代码：
#include<iostream>
using namespace std;
int main()
{
int N;
int *input;
int i;
int begin=0, end=0, sum=0; //最终所求的子序列的起始位置，终止位置，以及子序列和。
int tempSum=0, tempBegin=0, tempEnd=0; //目前正在考察的子序列的起始位置，终止位置，以及子序列的和。
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 {
//如果tempSum<0,那么tempSum+input[i]<input[i]
//所以此时我们要开始考察新的子序列
tempSum = 0;
tempSum += input[i];
tempBegin = i;
tempEnd = i;
}
//if(tempSum > sum) 这样写不能AC，应改为如下：
if(tempSum > sum || (tempSum == 0 && end == N-1))
{
sum = tempSum;
begin = tempBegin;
end = tempEnd;
}
}
cout<<sum<<" "<<input[begin]<<" "<<input[end]<<endl;
return 0;
}