1688 reverse order, 1688 Reverse Order

1688 reverse order, 1688 Reverse Order
1688 Reverse Order

 

Time Limit: 1 s space limit: 128000 KB title level: Gold Title Description Description

Given a sequence a1, a2 ,..., An. If I <j and ai> aj exists, we call it a reverse order pair to find the number of Reverse Order pairs.

 

Data range: N <= 105. Ai <= 105. The time limit is 1 s.

 

Input description Input Description

The first behavior n indicates the sequence length. The next n rows indicate the number of I in the sequence.

Output description Output Description

Total number of all reverse order pairs.

Sample Input Sample Input

4

3

2

3

2

Sample output Sample Output

3

Data range and prompt Data Size & Hint 

1 # include <iostream> 2 # include <cstring> 3 using namespace std; 4 long int a [1000001]; 5 long int tot; 6 long int n; 7 long int ans [1000001]; // storage result 8 long int now; // It indicates how many results 9 void f (long int s, long int t) 10 {11 // int mid, I, j, k; 12 if (s = t) return; 13 int mid = (s + t) /2; 14 f (s, mid); 15 f (mid + 1, t); 16 long int I = s; 17 long int j = mid + 1; 18 now = s; 19 while (I <= mid & j <= t) 20 {21 if (a [I] <= a [j]) 22 {23 // tot ++; 24 25 ans [now] = a [I]; 26 I ++; 27 now ++; 28} 29 else30 {31 tot = tot + mid-I + 1; 32 ans [now] = a [j]; 33 j ++; 34 now ++; 35} 36} 37 while (I <= mid) 38 {39 ans [now] = a [I]; 40 I ++; 41 now ++; 42} 43 while (j <= t) 44 {45 ans [now] = a [j]; 46 j ++; 47 now ++; 48} 49 for (I = s; I <= t; I ++) // put the merged ordered data back into array a 50 a [I] = ans [I]; 51 52} 53 int main () 54 {55 long int n; 56 cin> n; 57 for (int I = 1; I <= n; I ++) 58 cin> a [I]; 59 f (1, n ); 60 cout <tot; 61/* for (int I = 1; I <= n; I ++) 62 {63 cout <a [I] <""; 64} */65 return 0; 66}