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\n%d", ans, brr[0]);
64 for(int i = 1; i <= n; i++) {
65 printf(" %d", brr[i]);
66 }
67 printf("\n");
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。。。。
时间： 2024-10-07 06:49:28