01-复杂度1 最大子列和问题

01-复杂度1 最大子列和问题   (20分)给定KK个整数组成的序列{ N_1N​1​​, N_2N​2​​,..., N_KN​K​​ }，“连续子列”被定义为{ N_iN​i​​, N_{i+1}N​i+1​​,..., N_jN​j​​ }，其中 1\le i \le j \le K1≤i≤j≤K。“最大子列和”则被定义为所有连续子列元素的和中最大者。例如给定序列{-2, 11, -4, 13, -5, -2 }，其连续子列{ 11, -4, 13 }有最大的和20。现要求你编写程序，计算给定整数序列的最大子列和。本题旨在测试各种不同的算法在各种数据情况下的表现。各组测试数据特点如下：数据1：与样例等价，测试基本正确性；数据2：102个随机整数；数据3：103个随机整数；数据4：104个随机整数；数据5：105个随机整数；

输入格式:

输入第1行给出正整数KK (\le100000≤100000)；第2行给出KK个整数，其间以空格分隔。

输出格式:

在一行中输出最大子列和。如果序列中所有整数皆为负数，则输出0。

输入样例:

6
-2 11 -4 13 -5 -2

输出样例:

20

//求最大子列和三种方法用时比较//1.穷举法 2.在线处理 3.分而治之//the time complexity of the first is O(n2), the second is O(n), the third is O(nlgn).#include <iostream>#include <time.h>#define NUM 100using namespace std;int list[NUM];int compare(int a, int b, int c);int maxSequence1(int list[], int num);int maxSequence2(int list[], int num);int maxSequence3(int list[], int left, int right);clock_t start, end;double duration;int main(){int max;int num, n = 0;while(cin>>num){list= num;n++;}method_1:start = clock();for(int count = 0; count<=1e6; count++){max = maxSequence1(list, n);}cout<<"max1="<<max<<endl;end = clock();duration = ((double)(end - start))/CLK_TCK;cout<<"method 1 lasts "<<duration<<"s"<<endl;method_2:start = clock();for(int count = 0; count<=1e6; count++){max = maxSequence2(list, n);}cout<<"max2="<<max<<endl;end = clock();duration = ((double)(end - start))/CLK_TCK;cout<<"method 2 lasts "<<duration<<"s"<<endl;mehtod_3:start = clock();for(int count = 0; count<=1e6; count++){max = maxSequence3(list, 0, n);}cout<<"max3="<<max<<endl;end = clock();duration = ((double)(end - start))/CLK_TCK;cout<<"method 3 lasts "<<duration<<"s"<<endl;return 0;}int compare(int a, int b, int c){return a > b ? a > c ? a : c : b > c ? b : c;}int maxSequence1(int list[], int num){int max = 0;for(int i = 0; i < num; i++){int tempMax = 0;for(int j = i; j < num; j++){tempMax += list[j];if(tempMax > max){max = tempMax;}}}return max;}int maxSequence2(int list[], int num){int tempMax = 0;int max = 0;for(int i = 0; i < num; i++){tempMax += list[i];if(tempMax < 0){tempMax = 0;}else if(tempMax > max){max = tempMax;}}return max;}int maxSequence3(int list[], int left, int right){int leftMax, rightMax;if(left == right){// if(list[left] == 0){// return list[left];// }// else{// return 0;// }return list[left];}int middle;middle = (left + right) / 2;leftMax = maxSequence3(list, left, middle);rightMax = maxSequence3(list, middle+1, right);int tempLeft = 0, tempRight = 0;int leftBorderMax = 0, rightBorderMax = 0;for(int i = middle; i >= left; i--){tempLeft += list[i];if(tempLeft > leftBorderMax){leftBorderMax = tempLeft;}}for(int i = middle+1; i<= right; i++){tempRight += list[i];if(tempRight > rightBorderMax){rightBorderMax = tempRight;}}int middleMax = rightBorderMax + leftBorderMax;return compare( leftMax, rightMax, middleMax);}