01- 1- complexity of the problem and the biggest sub-columns - programming problem

Problem-solving Code

#include<stdio.h>

int MaxSubseqSum1(int a[],int n);
int MaxSubseqSum2(int a[],int n);
int MaxSubseqSum3(int a[],int n);
int MaxSubseqSum4(int a[],int n);

int main(void){
	int n;
	scanf("%d",&n);
	int a[n];
	int i;
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	int max;
//	max=MaxSubseqSum1(a,n);
//	max=MaxSubseqSum2(a,n);
//  max=MaxSubseqSum3(a,n);
	max=MaxSubseqSum4(a,n);
	printf("%d\n",max);
	return 0;
}

int MaxSubseqSum1(int a[],int n)
{
	int i,j,k;
	int sum;
	int max=0;
	for(i=0;i<n;i++){
		for(j=i;j<n;j++){
			sum=0;
			for(k=i;k<=j;k++){
				sum+=a[k];
			}
			if(sum>max){
				max=sum;
			}
		}
	}
	return max;
}

int MaxSubseqSum2(int a[],int n)
{
	int i,j;
	int sum;
	int max=0;
	for(i=0;i<n;i++){
		sum=0;
		for(j=i;j<n;j++){
			sum+=a[j];
			if(sum>max){
				max=sum;
			}
		}
	}
	return max;
}

int Max3( int A, int B, int C )
{ /* 返回3个整数中的最大值 */
    return A > B ? A > C ? A : C : B > C ? B : C;
}
 
int DivideAndConquer( int List[], int left, int right )
{ /* 分治法求List[left]到List[right]的最大子列和 */
    int MaxLeftSum, MaxRightSum; /* 存放左右子问题的解 */
    int MaxLeftBorderSum, MaxRightBorderSum; /*存放跨分界线的结果*/
 
    int LeftBorderSum, RightBorderSum;
    int center, i;
 
    if( left == right )  { /* 递归的终止条件，子列只有1个数字 */
        if( List[left] > 0 )  return List[left];
        else return 0;
    }
 
    /* 下面是"分"的过程 */
    center = ( left + right ) / 2; /* 找到中分点 */
    /* 递归求得两边子列的最大和 */
    MaxLeftSum = DivideAndConquer( List, left, center );
    MaxRightSum = DivideAndConquer( List, center+1, right );
 
    /* 下面求跨分界线的最大子列和 */
    MaxLeftBorderSum = 0; LeftBorderSum = 0;
    for( i=center; i>=left; i-- ) { /* 从中线向左扫描 */
        LeftBorderSum += List[i];
        if( LeftBorderSum > MaxLeftBorderSum )
            MaxLeftBorderSum = LeftBorderSum;
    } /* 左边扫描结束 */
 
    MaxRightBorderSum = 0; RightBorderSum = 0;
    for( i=center+1; i<=right; i++ ) { /* 从中线向右扫描 */
        RightBorderSum += List[i];
        if( RightBorderSum > MaxRightBorderSum )
            MaxRightBorderSum = RightBorderSum;
    } /* 右边扫描结束 */
 
    /* 下面返回"治"的结果 */
    return Max3( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum );
}
 
int MaxSubseqSum3( int a[], int n )
{ /* 保持与前2种算法相同的函数接口 */
    return DivideAndConquer( a, 0, n-1 );
}

int MaxSubseqSum4(int a[],int n)
{
	int i;
	int sum=0,max=0;
	for(i=0;i<n;i++){
		sum+=a[i];
		if(sum>max){
			max=sum;
		}
		if(sum<0){
			sum=0;
		}
	}
	return max;
}

Test Results

Finishing problem