int res = 0;
for(int i=0;i<A.length;i++){for(int j=i+1;j<A.length;j++){if(Fun(A[i],A[j]) == true) res += 1;}
}
return res;

/******* 利用字典树保存数组中所有的数*2的值，之后从右向左遍历数组，首先将这个数*2的值从树中删除，然后查找树中比* 这个数小的元素的个数。* 最坏时间复杂度: O(96 * n)* 插入、删除、查找还利用了尾递归来加快速度**/
public class Solution {public int reversePairs(int[] nums) {int res = 0,lenn = nums.length;//,halfMax = (Integer.MAX_VALUE)/2+10;if(lenn<2) return 0;Node head = new Node();long temp = nums[lenn-1];//if(nums[lenn-1] < halfMax) addNum(temp *2,head,63);for(int i=lenn-2;i>-1;i--){temp = nums[i];res += getRes(temp,head,63,0);addNum(temp *2,head,63);}return res;}private void delNum(long tar,Node he,int bit){// 删除树中这个元素，并且更新相关节点的值if(bit<0)return ;if((tar & (((long)1)<<bit)) == 0){if(he.zero.num > 0)he.zero.num -= 1;//else return ;he = he.zero;}else{if(he.one.num > 0)he.one.num -= 1;//else return ;he = he.one;}delNum(tar,he,bit-1);}private void addNum(long tar,Node he,int bit){// 向树中增加这个元素，并且更新相关节点的值if(bit<0)return;if((tar & (((long)1)<<bit)) == 0){if(he.zero == null) he.zero = new Node();else he.zero.num += 1;he = he.zero;}else{if(he.one == null) he.one = new Node();else he.one.num += 1;he = he.one;}addNum(tar,he,bit-1);}private int getRes(long tar,Node he,int bit,int res){// 符号位特殊判断if(tar>=0){if(he.one != null)res += he.one.num;he = he.zero;}else{he = he.one;}return getSmallerNum(tar,he,bit-1,res);}private int getSmallerNum(long tar,Node he,int bit,int res){// 获取比tar小的所有数的个数，据说是尾递归if(bit<0 || he == null) return res;if((tar & (((long)1)<<bit)) == 0){he = he.zero;}else{if(he.zero != null)res += he.zero.num;he = he.one;}return getSmallerNum(tar,he,bit-1,res);}private static class Node{public int num;public Node zero,one;public Node(){num = 1;zero = null;one = null;}}
}