第一周编程作业: Maximum Subsequence Sum

01-复杂度2 Maximum Subsequence Sum（25 point(s)） Given a sequence of K

integers { N ​1 ​​ , N ​2 ​​ , …, 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

程序实现如下：

#include <stdio.h>
#include <stdlib.h>
//void print_arr(int n, int * arr);

int main(void)
{
int n = 0;
scanf("%d", &n);
int * arr = (int *)malloc(n * sizeof(int));

for (int i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
getchar();
}

//int arr[5]={1,2,-8,2,1};
//n=5;

//n=10;
//int arr[10]={10,-2,5,5,-1,-100,100,-2,1,2};
//int arr[10]={-1,-2,-3,-4,-8,-3,-8,-9,-7,-11};//all the number are negative
//int arr[10]={-5,-2,8,-1,9,1,-10,0,0,0};//the sum have a negative number
//int arr[10]={};//all the number is zero
//int arr[10]={0,0,10,-1,-1,5,-1,-1,-1,-100};
//int arr[10]={-2,-5,1,50,-1,-1,-2,-300,-4,-8};
//int arr[10]={100,-8,-8,1,-8,-1000,1,2,0,0};

//int arr[10]={-1,-2,-3,-5,-1,0,-1,-1,-1,-2};

int next_sum, current_sum;
next_sum = current_sum = 0;

int start=0, end=n-1;//final result
int start1=0,end1=n-1;//maybe the final result
int flag=0;
int tempSum=0;
int s=0;
int j=0;
int k=1;
int t=0;

// int flag2 = 0;
//int start_temp=0;

for (int i = 0; i < n; i++)//check all the numbers
{
current_sum += arr[i];
if (flag == 0)
{
if (arr[i] > 0|| arr[i]==0)//find a posstive number
{
start1 = i;//mark it
t=start1;
//start=start1;
//printf("**********%d\n",start1);
if(arr[i]==0)
{
k++;
}
flag = 1;
}
else//can't find a suitable number
{
current_sum=0;
//printf("positive!!\n");
k++;
continue;

}
}
if ( current_sum < 0)
{
s=current_sum;
tempSum=current_sum-arr[i];//maybe the maxsum
start1=i+1;
j=i-1;
while(arr[j]<0&&s==tempSum)
{
s=s-arr[j];
end=j-1;
//start=start1;
j--;
// printf("*******%d\n",end);
}

if(tempSum<current_sum)//when it's larger than before
{

start=start1;
end=end1;
}

current_sum = 0;
flag=0;

}
else if (current_sum > next_sum)
{
next_sum = current_sum;
end1 = i;
// printf("*******end1=%d\n",end1);
if(tempSum<current_sum)//when it's larger than before
{
start=start1;
end=end1;
}
}

}
if(k==n)
{
start=t;
end=t;
}

// printf("less than max=%d\n",tempSum);
// printf("sum=%d %d %d\n", next_sum,start,end);
//    for(int i=start;i<stop+1;i++)
//    {

//        next_sum=next_sum+arr[i];
//    }
printf("%d %d %d", next_sum, arr[start], arr[end]);
//print_arr(n, arr );

return 0;
}