这个题一开始实在是没好办法, 主要是没法确定最大子列的第一项,由于整个序列的最大子列的最后一项很好确定,于是乎,我把整个序列倒过来再求一次最大子列的最后一项,即为整个序列正序的第一项。#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
scanf("%d", &n);
int i,temp =1;
int a[100000];
for(i = 0;i < n; i++)
{
scanf("%d", &a[i]);
}
int thissum, maxsum,maxsum2;
int p1,p2;
thissum = maxsum = maxsum2=0;
for(i = 0; i < n ; i++)
{
thissum += a[i];
if(thissum > maxsum)
{
maxsum = thissum;
p2 = a[i];//Ëø¶¨Ä©Î²µÄλÖÃ
}
else if(thissum < 0)
thissum = 0;
}
thissum = maxsum = 0;
for(i = n-1; i >= 0 ; i--)
{
thissum += a[i];
if(thissum >= maxsum)
{
maxsum = thissum;
p1 = a[i];
}
else if(thissum < 0)
thissum = 0;
if(a[i] >= 0)
temp = 0;
}
if(temp)
printf("%d %d %d\n", maxsum,a[0],a[n-1]);
else
printf("%d %d %d\n", maxsum,p1,p2);
return 0;
}
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 -21
Sample Output:
10 1 4
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
scanf("%d", &n);
int i,temp =1;
int a[100000];
for(i = 0;i < n; i++)
{
scanf("%d", &a[i]);
}
int thissum, maxsum,maxsum2;
int p1,p2;
thissum = maxsum = maxsum2=0;
for(i = 0; i < n ; i++)
{
thissum += a[i];
if(thissum > maxsum)
{
maxsum = thissum;
p2 = a[i];//Ëø¶¨Ä©Î²µÄλÖÃ
}
else if(thissum < 0)
thissum = 0;
}
thissum = maxsum = 0;
for(i = n-1; i >= 0 ; i--)
{
thissum += a[i];
if(thissum >= maxsum)
{
maxsum = thissum;
p1 = a[i];
}
else if(thissum < 0)
thissum = 0;
if(a[i] >= 0)
temp = 0;
}
if(temp)
printf("%d %d %d\n", maxsum,a[0],a[n-1]);
else
printf("%d %d %d\n", maxsum,p1,p2);
return 0;
}
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 -21
Sample Output:
10 1 4