【挑戰程式設計競賽】分治法求一個數列逆序對的對數

#include <iostream>#include <vector>using namespace std;typedef long long LL;LL merge_cnt(vector<int> &input){int n = input.size();if(n <= 1) return 0;LL cnt = 0;vector<int> left(input.begin(),input.begin() + n/2); //劃分 vector<int> right(input.begin() + n/2,input.end());cnt += merge_cnt(left);cnt += merge_cnt(right);int ai = 0, bi = 0, ci = 0;while(ai < n){if(bi < left.size() && (ci == right.size() || left[bi] <= right[ci])){input[ai++] = left[bi++];}else{cnt += n/2 - bi;         // 對於數列right的每一個元素，統計left中 // 大於此元素的個數，此時兩個數列是已經排好序的 input[ai++] = right[ci++];}}return cnt;} int main(){int a[4]={3,1,4,2};vector<int> input(a,a+4);cout<<merge_cnt(input)<<endl;for(int i = 0; i < input.size(); i++){cout<<input[i]<<" ";}cout<<endl;}