Maximum sub columns and problems

Http://www.patest.cn/contests/mooc-ds/01-1

Given the sequence of K-integers {N1, N2, ..., NK}, "contiguous sub columns" are defined as {Ni, ni+1, ..., Nj}, where 1 <= i <= J <= K. Maximum sub columns and is defined as the largest of all contiguous child column elements. For example, given the sequence {-2, 11,-4, 13,-5,-2}, its contiguous sub columns {11,-4, 13} have the largest and 20. You are asked to write a program that calculates the maximum number of columns for a given integer sequence.

Input format:

The input line 1th gives a positive integer k (<= 100000), and line 2nd gives K integers, which are separated by spaces.

Output format:

Outputs the largest columns in a row. If all integers in the sequence are negative, the output is 0. Input Sample:

6
-2 11-4 13-5-2

Output Sample:

20

O (n*3) Solution:

#include <stdio.h>
#define MAXN 100001
Long Long maxSubseqSum1 (int a[], int n) {
	long long maxsum = 0;
  for (int i = 0; i < n; ++i) {for
		(int j = i; j < n; ++j) {
			long long thissum = 0;
			for (int k = i; k <= J; ++k) {
				thissum = a[k];
				if (Thissum > Maxsum)
					maxsum = thissum;
			}
	}} return maxsum;
}

int main () {
	int k, A[MAXN];
	scanf ("%d", &k);
	for (int i = 0; i < K; ++i)
		scanf ("%d", &a[i]);
	printf ("%lld\n", MaxSubseqSum1 (A, k));
	return 0;
}

O (n*2) Solution:

#include <stdio.h>
#define MAXN 100001
Long Long maxSubseqSum2 (int a[], int n) {
	long long maxsum = 0;
  for (int i = 0; i < n; ++i) {
		long long thissum = 0;
		for (int j = i; j <= N; ++j) {
			thissum = a[j];
			if (Thissum > Maxsum)
				maxsum = thissum;
			if (Thissum < 0)
				thissum = 0;
		}
	}
	return maxsum;
}

int main () {
	int k, A[MAXN];
	scanf ("%d", &k);
	for (int i = 0; i < K; ++i)
		scanf ("%d", &a[i]);
	printf ("%lld\n", maxSubseqSum2 (A, k));
	return 0;
}

Divide-and-conquer algorithm O (NLOGN) Solution:

#include <stdio.h> #include <vector> #include <iostream> using namespace std;  
    Long Long Max (long A, long long B, long long C) {if (b > a) a = B;  
    if (a > C) return A;  
else return C; Long Long maxsubseqsum (const vector<int> &a, int left, int right) {if (left = right) {if (a[l  
        EFT] > 0) return a[left];  
    else return 0;  
    int mid = (left + right)/2;  
    Long Long maxleftsum = Maxsubseqsum (A, left, mid);  
    Long Long maxrightsum = Maxsubseqsum (A, mid+1, right);  
    Long Long maxleftbordersum = 0, leftbordersum = 0;  
        for (int i = mid; I >= left;-i) {leftbordersum + = A[i];  
    if (Leftbordersum > maxleftbordersum) maxleftbordersum = leftbordersum;  
    Long Long maxrightbordersum = 0, rightbordersum = 0; for (int i = mid+1 i <= right; ++i) {Rightbordersum + + a[i];  
    if (Rightbordersum > maxrightbordersum) maxrightbordersum = rightbordersum;  
Return Max (Maxleftsum, Maxrightsum, maxleftbordersum+maxrightbordersum);  
    int main () {vector<int> A;  
    int k, x;  
    scanf ("%d", &k);  
        for (int i = 0; i < K; ++i) {cin >> x;  
    A.push_back (x);  
    printf ("%lld\n", Maxsubseqsum (A, 0, a.size ()-1));  
return 0;   }

Linear algorithm O (n) Complexity:

#include <stdio.h>
#define MAXN 100001
Long Long maxsubseqsum (int a[], int n) {
	long long thissum = 0, MA xsum = 0;
	for (int i = 0; i < n; ++i) {
		thissum + = A[i];
		if (Thissum > Maxsum)
			maxsum = thissum;
		else if (Thissum < 0)
			thissum = 0;
	}
	return maxsum;
}

int main () {
	int k, A[MAXN];
	scanf ("%d", &k);
	for (int i = 0; i < K; ++i)
		scanf ("%d", &a[i]);
	printf ("%lld\n", Maxsubseqsum (A, k));
	return 0;
}