D. Kuro and GCD and XOR and SUM

Kuro is currently playing an educational game about numbers. The game focuses on the greatest common divisor (GCD), the XOR value, and the sum of two numbers. Kuro loves the game so much that he solves levels by levels day by day.

Sadly, he's going on a vacation for a day, and he isn't able to continue his solving streak on his own. As Katie is a reliable person, Kuro kindly asked her to come to his house on this day to play the game for him.

Initally, there is an empty array $\mathrm{aa.\; The\; game\; consists\; of qq tasks\; of\; two\; types.\; The\; first\; type\; asks\; Katie\; to\; add\; a\; number uiui to aa.\; The\; second\; type\; asks\; Katie\; to\; find\; a\; number vv existing\; in aa such\; that ki\mid GCD(xi,v)ki\mid GCD(xi,v), xi+v\le sixi+v\le si,\; and xi\oplus vxi\oplus v is\; maximized,\; where \oplus \oplus denotes\; the bitwise\; XOR\; operation, GCD(c,d)GCD(c,d) denotes\; the greatest\; common\; divisor of\; integers cc and dd,\; and y\mid xy\mid x means xx is\; divisible\; by yy,\; or\; report -1 if\; no\; such\; numbers\; are\; found.}$

Since you are a programmer, Katie needs you to automatically and accurately perform the tasks in the game to satisfy her dear friend Kuro. Let's help her!

Input

The first line contains one integer $\mathrm{qq (2\le q\le 1052\le q\le 105)\; —\; the\; number\; of\; tasks\; the\; game\; wants\; you\; to\; perform.}$

$\mathrm{qq lines\; follow,\; each\; line\; begins\; with\; an\; integer titi —\; the\; type\; of\; the\; task:}$

• If ${\mathrm{ti=1ti=1,\; an\; integer uiui follow\; (1\le ui\le 1051\le ui\le 105)\; —\; you\; have\; to\; add uiui to\; the\; array aa.}}^{}$
• If ${\mathrm{ti=2ti=2,\; three\; integers xixi, kiki,\; and sisi follow\; (1\le xi,ki,si\le 1051\le xi,ki,si\le 105)\; —\; you\; must\; find\; a\; number vv existing\; in\; the\; array aa such\; that ki\mid GCD(xi,v)ki\mid GCD(xi,v), xi+v\le sixi+v\le si,\; and xi\oplus vxi\oplus v is\; maximized,\; where \oplus \oplus denotes\; the\; XOR\; operation,\; or\; report -1 if\; no\; such\; numbers\; are\; found.}}^{}$

It is guaranteed that the type of the first task is type $\mathrm{11,\; and\; there\; exists\; at\; least\; one\; task\; of\; type 22.}$

Output

For each task of type $\mathrm{22,\; output\; on\; one\; line\; the\; desired\; number vv,\; or -1 if\; no\; such\; numbers\; are\; found.}$

Examples

input

5
1 1
1 2
2 1 1 3
2 1 1 2
2 1 1 1

output

2
1
-1

input

10
1 9
2 9 9 22
2 3 3 18
1 25
2 9 9 20
2 25 25 14
1 20
2 26 26 3
1 14
2 20 20 9

output

9
9
9
-1
-1
-1

Note

In the first example, there are 5 tasks:

• The first task requires you to add $\mathrm{11 into aa. aa is\; now \{1\}\{1\}.}$
• The second task requires you to add $\mathrm{22 into aa. aa is\; now \{1,2\}\{1,2\}.}$
• The third task asks you a question with $\mathrm{x=1x=1, k=1k=1 and s=3s=3.\; Taking\; both 11 and 22 as vv satisfies 1\mid GCD(1,v)1\mid GCD(1,v) and 1+v\le 31+v\le 3.\; Because 2\oplus 1=3>1\oplus 1=02\oplus 1=3>1\oplus 1=0, 22 is\; the\; answer\; to\; this\; task.}$
• The fourth task asks you a question with $\mathrm{x=1x=1, k=1k=1 and s=2s=2.\; Only v=1v=1 satisfies 1\mid GCD(1,v)1\mid GCD(1,v) and 1+v\le 21+v\le 2,\; so 11 is\; the\; answer\; to\; this\; task.}$
• The fifth task asks you a question with $\mathrm{x=1x=1, k=1k=1 and s=1s=1.\; There\; are\; no\; elements\; in aa that\; satisfy\; the\; conditions,\; so\; we\; report-1 as\; the\; answer\; to\; this\; task.}$

这题其实可以说是01字典树的裸题，  看了别人的题解 然后就去学了一下01字典树

现在我也只是刚刚学有点懵逼

    我看这篇博客学的 转载

   源地址https://blog.csdn.net/riba2534/article/details/80344026

     

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<vector>
  5 #include<queue>
  6 #define inf 0x3fffffff
  7 
  8 using namespace std;
  9 typedef long long LL;
 10 const int maxn = 2e5 + 10;
 11 int a[maxn], minn[maxn], tree[32 * maxn][2];
 12 int sum, root;
 13 int newnode() {
 14     tree[sum][0] = tree[sum][1] = -1;
 15     sum++;
 16     return sum - 1;
 17 }
 18 
 19 void init() {
 20     sum = 0 ;
 21     memset(minn, 0x3f, sizeof(minn));
 22     root = newnode();
 23 }
 24 
 25 /*
 26 void add(int x) {
 27     int p = x;
 28     int now = root;
 29     minn[now] = min(minn[now], p);
 30     for (int i = 31 ; i >= 0 ; i--) {
 31         int to = (x >> i ) & 1;
 32         if (tree[now][to] == -1)  tree[now][to] = newnode();
 33         now = tree[now][to];
 34         minn[now] = min(minn[now], p);
 35     }
 36 }*/
 37 void add(int x) {
 38     int p = x;
 39     int now = root ;
 40     minn[now] = min(minn[now], p);
 41     for (int i = 31 ; i >= 0 ; i--) {
 42         int  to = (x >> i) & 1;
 43         if (tree[now][to] == -1)  tree[now][to] = newnode();
 44         now = tree[now][to];
 45         minn[now] = min(minn[now], p);
 46     }
 47 }
 48 /*
 49 int getans(int x, int p) {
 50     int now = root;
 51     if (minn[now] > p) return -1;
 52     for (int i = 31 ; i >= 0 ; i-- ) {
 53         int to = (x >> i) & 1 ;
 54         if (tree[now][to ^ 1] != -1 && minn[tree[now][to ^ 1]] <= p   ) to ^= 1;
 55         now = tree[now][to];
 56     }
 57     return minn[now];
 58 }*/
 59 int getans(int x, int p) {
 60     int now = root ;
 61     if (minn[now] > p ) return -1;
 62     for (int i = 31 ; i >= 0 ; i--  ) {
 63         int to = (x >> i) & 1;
 64         if (tree[now][to ^ 1] != -1 && minn[tree[now][to ^ 1]] <= p ) to ^= 1;
 65         now = tree[now][to];
 66     }
 67     return minn[now];
 68 }
 69 int main() {
 70     int t, op, x, k, s;
 71     init();
 72     scanf("%d", &t);
 73     while(t--) {
 74         scanf("%d", &op);
 75         if (op == 1) {
 76             scanf("%d", &x);
 77             a[x] = 1;
 78             add(x);
 79         } else {
 80             scanf("%d%d%d", &x, &k, &s);
 81             if (x % k != 0) {
 82                 printf("-1\n");
 83                 continue;
 84             }
 85             if (k == 1) {
 86                 printf("%d\n", getans(x, s - x));
 87                 continue;
 88             }
 89             int maxn = -1, ans = -1;
 90             for (int i = k ; i <= s - x ; i += k)  {
 91                 if (a[i] && (i ^ x) > maxn) {
 92                     maxn = i ^ x;
 93                     ans = i;
 94                 }
 95             }
 96             printf("%d\n", ans);
 97         }
 98     }
 99     return 0;
100 }