分治法之求最大连续子序列和

【说人话的算法小课堂】使用分治法求最大连续子序列的和 使用分治法求最大连续子序列的和 设有一个整数数列a[n],1≤s, t≤n。求a[n]的一个子数列a[s, t],使得该子数列的和尽可能大。 若这个和小于0,则输出0。 【算法本体】 (参考代码中,因为要与暴力法对比运行时间,而暴力法用到前缀和,为了方便,数组下标从1开始) 设m = floor((s + t) / 2)。 当n = 1时,直接取唯一的元素作为结果。若该元素为负,则输出0。 当n > 1时,分3种情况讨论: 【情况1】该子数列完全落在原数列的左半部分a[s, m]。递归对数列a[s, 阅读详情

对原问题有如下解
1.最大子序列在数组中点的左边
2.最大子序列在数组中点的右边
3.最大子序列跨越数组中点

#include<iostream>
using namespace std;
int FIND_MAX_CROSSING_SUBARRAY(int a[],int low,int high)
{
    int mid = (low+high)/2;
    int i = mid;
    int j = mid+1;
    int left_sum = 0;
    int right_sum = 0;
    int sum = 0;
    while(i>=low){
     
对求最大连续子序列和问题四种算法的分析 对求最大连续子序列和问题四种算法的分析(暴力枚举、优化枚举、动态规划、分治递归) 阅读详情

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