输入一个数值序列numbers，输出某一个连续子串长度，满足小于给定乘积上限k。
这里有两种解法，其一为O（N^2）,其二基本为O（N）
1 #include<iostream> 2 #include <stdio.h> 3 #include<algorithm> 4 #include<vector> 5 #include<map> 6 7 using namespace std; 8 9 long getSubArray(vector<int> &numbers, int k) 10 { 11 long count = 0; 12 for (int i = 0; i < numbers.size();++i) { 13 long mul = numbers[i]; 14 if (mul <= k) { 15 count++; 16 for (int j = i+1; j < numbers.size();++j) { 17 mul *= numbers[j]; 18 if (mul <= k) { 19 count++; 20 } 21 else{ 22 break; 23 } 24 } 25 } 26 } 27 return count; 28 } 29 30 long getSubArray2(vector<int> &numbers, int k) 31 { 32 vector<int> loc(numbers.size(), 0); 33 vector<long> muls(numbers.size(), 0); 34 int i = 0; 35 int mul = numbers[i]; 36 if (mul <= k) { 37 int j = i + 1; 38 for (;j < numbers.size();++j) { 39 mul *= numbers[j]; 40 if (mul > k) { 41 break; 42 } 43 } 44 loc[i] = j; 45 muls[i] = mul; 46 } 47 else 48 { 49 loc[i] = i; 50 muls[i] = numbers[i]; 51 } 52 53 for (int i = 1;i < numbers.size();++i) { 54 int mul = muls[i-1]/(numbers[i-1]); 55 if (mul <= k) { 56 int j = loc[i-1] + 1; 57 for (; j < numbers.size(); j++){ 58 mul *= numbers[j]; 59 if (mul > k) { 60 break; 61 } 62 } 63 loc[i] = j; 64 muls[i] = mul; 65 } 66 else{ 67 loc[i] = loc[i - 1] > i ? loc[i - 1] : i; 68 muls[i] = mul; 69 } 70 } 71 72 long count = 0; 73 for (int i = 0; i < numbers.size();++i) { 74 count += loc[i] - i; 75 } 76 return count; 77 } 78 79 80 int main() 81 { 82 cout << "hello, world!" << endl; 83 int i = 5, j = 2; 84 int k=i<<j; 85 printf("%d %d ", 0xff , k); 86 printf("%d %d ", i << j, j >> i); 87 88 vector<int> arr = {2,34,11,1,2,3,4,5,6,65,33,1,1,23,2,4,2,123,24,12,2,22}; 89 int m = 20; 90 int count = getSubArray(arr, m); 91 int count2 = getSubArray2(arr, m); 92 cout << count <<" "<<count2<< endl; 93 system("pause"); 94 return 0; 95 }