//题意:此题题意,给定一个顶长的序列,问每次只能反转相邻的两个数字,问最少次数能让这个序列排成从小到大的顺序,说白了就是求逆序数。如:8 6 0 2 4即:4+3=7。若是求从大到小的顺序求最少次的反转如:8 6 0 2 4即:2+1=3(逆序数是找从当前下标向后比当前元素小的元素个数,若求从大到小那么就是 找比当前大的个数,自行体会一下)
//思路:用归并排序,对同一区间划分左右两个区间,若左区间的元素大于右区间的元素,则len_r-i+1即可,详情见代码
#include <iostream> #include <cstdio> #define maxx 1000000000 using namespace std; __int64 cnt; void cal(int *n, int l, int mid, int r) { int len_l = mid - l + 1; int len_r = r - mid; int *left = new int[len_l+2]; int *right = new int[len_r+2]; left[len_l+1] = maxx; //这里需要注意的当某个区间归并完后,另一个区间剩余元素直接相当于copy所以,给定越界值为最大值 right[len_r+1] = maxx; //同上 int i, j; i = j = 1; for(; i <= len_l; i++) left[i] = n[l+i-1]; for(; j <= len_r; j++) right[j] = n[mid+j]; i = j = 1; for(int k = l; k <= r; ){ if(left[i] <= right[j]){ n[k++] = left[i++]; }else{ n[k++] = right[j++]; cnt = cnt + len_l - i + 1;//主要在这里计算逆序数,自行体会一下 } } delete left; delete right; return ; } void mergesort(int *n, int l, int r) { if(l < r){ int mid = (l+r)/2; mergesort(n, l, mid); mergesort(n, mid+1, r); cal(n, l, mid, r); } return ; } int main() { int t; while(scanf("%d", &t) != EOF && t){ cnt = 0; int *n = new int[t+1]; n[0] = maxx; for(int i = 1; i < t+1; i++){ scanf("%d", &n[i]); } mergesort(n, 1, t); printf("%I64d\n", cnt); delete n; } return 0; }