Maximum continuous product substring

1/* 2 A = ",-,-3", to obtain the maximum continuous product substring, there are three methods, Method 1: using dynamic programming method, it is the easiest to understand and implement. Method 2: Use the three ideas of dynamic planning, but there is no need to save two array spaces. Method 3: Use the Record maximum value, minimum value method 4 */5 6/* 7 dynamic planning method, two arrays, maximum and minimum, max [I] is the maximum continuous product of the sequence ending with I. 8 maxnum [I] = max {maxnum [I-1] * A [I], minnum [I-1] * A [I], a [I]}; 9 minnum [I] = min {maxnum [I-1] * A [I], minnum [I-1] * A [I], a [I]}; 10 maxnum [0] = minnum [0] = A [0]; 11 */12 # include <iostream> 13 using namespace STD; 14 # include <string> 15 # include <algorithm> 16 double maxmultiply (const double * STR, int N) 17 {18 int size = N; 19 if (size = 0) 20 return 0; 21 double * maxnum = new double [size]; 22 if (maxnum = NULL) 23 exit (1); 24 double * Minnum = new double [size]; 25 if (minnum = NULL) 26 exit (1); 27 double result = STR [0]; 28 maxnum [0] = minnum [0] = STR [0]; 29 for (INT I = 1; I <n; I ++) 30 {31 maxnum [I] = max (maxnum [I-1] * STR [I], minnum [I-1] * STR [I]), STR [I]); 32 minnum [I] = min (maxnum [I-1] * STR [I], minnum [I-1] * STR [I]), STR [I]); 33 If (maxnum [I]> result) 34 result = maxnum [I]; 35} 36 Delete [] maxnum; 37 Delete [] minnum; 38 return result; 39} 40 41/* 42 Dynamic Planning But two variables are used to save the max and Min sequences. 43 */44 double maxmultiply2 (const double * a, int N) 45 {46 if (a = NULL) 47 return 0; 48 double result = A [0]; 49 double maxnum = A [0], minnum = A [0]; 50 for (INT I = 1; I <n; I ++) 51 {52 double tmpmax = maxnum * A [I]; 53 double tmpmin = minnum * A [I]; 54 maxnum = max (tmpmax, tmpmin ), A [I]); 55 minnum = min (tmpmax, tmpmin), a [I]); 56 If (maxnum> result) 57 result = maxnum; 58} 59 return result; 60} 61 62 // method 3, using methods similar to observation and induction, maxcurrent, mincurrent, maxproduct, and minproduct. Note that if maxcurrent is less than 1, in fact, it is better to compare 63 // with yourself, but it is still the same, there is no need to write 64 double maxmultiply3 (const double * a, int N) 65 {66 if (a = NULL) 67 return 0; 68 double maxcurrent = 1; 69 double mincurrent = 1; 70 double maxproduct = 1; 71 double minproduct = 1; 72 for (INT I = 0; I <N; I ++) 73 {74 maxcurrent * = A [I]; 75 mincurrent * = A [I]; 76 if (maxcurrent> maxproduct) 77 maxproduct = maxcurrent; 78 If (mincurrent> maxproduct) 79 maxproduct = mincurrent; 80 If (maxcurrent <minproduct) 81 minproduct = maxcurrent; 82 If (mincurrent <minproduct) 83 minproduct = mincurrent; 84 If (mincurrent> maxcurrent) 85 swap (mincurrent, maxcurrent); 86 If (maxcurrent <1) 87 maxcurrent = 1; 88} 89 return maxproduct; 90} 91 92 int main () 93 {94 double A [] = {-2.5, 0.5, 3, 8,-1}; 95 int n = 7; 96 cout <maxmultiply3 (A, n) <Endl; 97 system ("pause"); 98}