1 // you can use includes, for example: 2 // #include <algorithm> 3  4 // you can write to stdout for debugging purposes, e.g. 5 // cout << "this is a debug message" << endl; 6  7 int solution(vector<int> &A) { 8     // write your code in C++11 9     int i = A.size()-1;10     int tmp = 0;11     int res = 0;12     while(i >= 0)13     {14         if(A[i] == 0)15         {16             res += tmp;17         }18         else19         {20             tmp++;21         }22         --i;23     }24     return res;25 }

思路：统计0后面的1的个数，利用后缀数组计算

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

• 0 represents a car traveling east,
• 1 represents a car traveling west.

The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

  A[0] = 0  A[1] = 1  A[2] = 0  A[3] = 1  A[4] = 1

We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A of N integers, returns the number of passing cars.

The function should return −1 if the number of passing cars exceeds 1,000,000,000.

For example, given:

  A[0] = 0  A[1] = 1  A[2] = 0  A[3] = 1  A[4] = 1

the function should return 5, as explained above.

Assume that:

• N is an integer within the range [1..100,000];
• each element of array A is an integer that can have one of the following values: 0, 1.

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Colidity--PassingCars