出入算法

1.最大子列和问题

1.二次循环，复杂度n^2

#include<stdio.h>
#define NU 100000
int findson(int a, int b[])
{
	int thissum,msum;
	msum = 0;
	for (int i = 0; i < a; i++)
	{
		thissum = 0;
		for (int j = i; j < a; j++)
		{
			thissum += b[j];
			if (thissum > msum)
				{
					msum = thissum;
				}
		}
	}
	return msum;
}
int main()
{
	int a;
	scanf("%d", &a);
	int b[NU];
	for (int i = 0; i < a; i++)
	{
		scanf("%d", &b[i]);
	}
	printf("%d",findson(a,b));
	return 0;
}

2.搜到的大佬的答案复杂度o=n
惊了！

int MaxSubseqSum4( int A[], int N ) {
    int ThisSum, MaxSum, i;
    ThisSum = MaxSum = 0;
    for( i = 0; i < N; i++ ) {
          ThisSum += A[i]; /* 向右累加 */
          if( ThisSum > MaxSum )
                  MaxSum = ThisSum; /* ·发现更大和则更新当前结果 */
          else if( ThisSum < 0 ) /* 如果当前子列和为负数 */
                  ThisSum = 0; /* 则不可能使后面的部分和增大，抛弃之 */
    }
    return MaxSum;  
}

3.分而治之T(n) = O(nlogn)
精了

int Max(int A, int B, int C) {
	return (A > B) ? (A > C ? A : C) : (B > C ? B : C);
}
int divideandbonder(int list[], int left, int right)
{
	if (left == right)
	{
		if (list[left] > 0)
			return list[left];
		else
		{
			return 0;
		}
	}
	int mmleftsum, maleftsum;
	int mmrightsum, marightsum,center;
	center = (left + right) / 2;
	maleftsum = 0; mmleftsum = 0; mmrightsum = 0; marightsum = 0;
	mmleftsum = divideandbonder(list, left, center);
	mmrightsum = divideandbonder(list, center + 1, right);
	for (int i = 0;  i < left; i++)
	{
		maleftsum += list[i];
			if (maleftsum > mmleftsum)
			{
				mmleftsum = maleftsum;
		}
	}
	for (int i = center + 1; i < right; i++)
	{
		marightsum += list[i];
		if (marightsum > mmrightsum)
		{
			mmrightsum = marightsum;
		}
	}
	return Max(mmleftsum, mmrightsum, mmleftsum + mmrightsum);
}
int findson(int a, int b[])
{
	return divideandbonder(b, 0, a);
}

之后还会更