思路：既然是求两个数的异或运算之和，且由于数字不重复，那么肯定两个数异或的结果数字越大越好，即异或后从ai二进制的最高位后全是1。
具体思路看代码：
AC代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #include <vector> 5 #include <algorithm> 6 using namespace std; 7 int bit[17]; 8 int brr[100005]; 9 int n; 10 void init() 11 { 12 bit[0] = 1; 13 for(int i = 0; i <= 16; i++) { 14 bit[i+1] = 2<<i; 15 } 16 } 17 int arr[100005]; 18 void deal(int x) 19 { 20 for(int i = 0; i < n+10; i++) arr[i] = -1; 21 arr[0] = 0; 22 int k, le, xx; 23 while(x != 0){ 24 if(x <= 0) return; 25 int pos = upper_bound(bit, bit+18, x) - bit - 1; 26 if(pos == 0) { 27 arr[0] = 1; arr[1] = 0; 28 return; 29 } 30 xx = x; 31 if(pos == 1) k = 0x00000003 ^ x; 32 else if(pos == 2) k = 0x00000007 ^ x; 33 else if(pos == 3) k = 0x0000000f ^ x; 34 else if(pos == 4) k = 0x0000001f ^ x; 35 else if(pos == 5) k = 0x0000003f ^ x; 36 else if(pos == 6) k = 0x0000007f ^ x; 37 else if(pos == 7) k = 0x000000ff ^ x; 38 else if(pos == 8) k = 0x000001ff ^ x; 39 else if(pos == 9) k = 0x000003ff ^ x; 40 else if(pos == 10) k = 0x000007ff ^ x; 41 else if(pos == 11) k = 0x00000fff ^ x; 42 else if(pos == 12) k = 0x00001fff ^ x; 43 else if(pos == 13) k = 0x00003fff ^ x; 44 else if(pos == 14) k = 0x00007fff ^ x; 45 else if(pos == 15) k = 0x0000ffff ^ x; 46 else if(pos == 16) k = 0x0001ffff ^ x; 47 le = k; 48 x = k - 1; 49 for(int i = xx; i >= k; i--) arr[i] = le++; 50 } 51 } 52 int main() 53 { 54 init(); 55 while(scanf("%d", &n) != EOF) { 56 deal(n); 57 long long ans = 0; 58 for(int i = 0; i <= n; i++) { 59 int a; scanf("%d", &a); 60 brr[i] = arr[a]; 61 ans += a ^ arr[a]; 62 } 63 printf("%I64d %d", ans, brr[0]); 64 for(int i = 1; i <= n; i++) { 65 printf(" %d", brr[i]); 66 } 67 printf(" "); 68 } 69 return 0; 70 }
总结：QAQ比赛的时候看错题意了。。。。忽略了数字不重复的条件，导致想了好久。。。。