Maximum subsequence and (continuous): Maximum Sequence

Maximum subsequence and (continuous): Maximum Sequence

// Maximum subsequence and (continuous): // http://my.oschina.net/itblog/blog/267860#include <iostream> using namespace std; int MaxSum (int * a, int n) {int sum = 0; int max = 0; // the first element of the largest subsequence cannot be a negative number, // It is impossible to include subsequences with the sum of 0 or negative numbers as the prefix. // This avoids multiple considerations for the same element for (int I = 0; I <n; I ++) {sum + = a [I]; if (sum> max) max = sum; else if (sum <0) sum = 0;} return sum ;} int main () {int a [] = {-1,-2,-3,-4}; // test case cout where all values are negative <MaxSum (a, 4) <endl; int B [10] = {1 ,- 2, 3, 10,-4, 7, 2,-5}; cout <MaxSum (B, 8) <endl; system ("pause "); return 0;}/* For example, array: 1,-2, 3, 10,-4, 7, 2, -5 the maximum subsequence is 13. One is the violent enumeration O (n ^ 3). Two for loops determine the boundary, and the third for loop traverses and sums up and compares. For (I = 0, I --- n) for (j = I, j --- n) for (k = I, k --- n) sum + = s [k] A traversal O (n ^ 2): In the second for loop, j moves one side and adds and then compares. For (I = 0, I --- n) for (j = I, j --- n) sum + = s [j] one is to use DP for consideration, the largest subsequence is either in the left half, or in the right half, or across the Left and Right O (nlogn ). One is linear, such as O (n )*/

// Divide and conquer method: // Let's look at recursion and binary # include <iostream> using namespace std; // calculate the maximum number of three int max3 (int I, int j, int k) {if (I> = j & I> = k) return I; return max3 (j, k, I);} int maxsequence2 (int a [], int l, int u) {if (l> u) return 0; if (l = u) return a [l]; int m = (l + u)/2; // calculate the left half part of the largest continuous subsequence across the Left and Right int lmax = a [m], lsum = 0; // This cycle is the maximum value of this sequence, add all elements for (int I = m; I> = l; I --) {lsum + = a [I]; if (lsum> lmax) lmax = lsum ;} // calculate the cross Int rmax = a [m + 1], rsum = 0; for (int I = m + 1; I <= u; I ++) {rsum + = a [I]; if (rsum> rmax) rmax = rsum;} // if the maximum subsequence spans the left and right half sides, that is, the sum of lmax on the left half side and rmax on the right half side. Return max3 (lmax + rmax, maxsequence2 (a, l, m), maxsequence2 (a, m + 1, u);} int main () {// int a [] = {-1,-2,-3,-4}; // test case where all values are negative. // cout <MaxSum (a, 4) <endl; int a [10] = {1,-2, 3, 10,-4, 7, 2,-5}; cout <maxsequence2 (a, 0, 8) <endl; system ("pause"); return 0 ;}