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Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

给出N
for(i = 1; i <= N; i ++) a[i] = i;
solve(n,a[]) {
     if(n == 1) return;
    将奇数下标的数赋给a1，假设有len1个
    将偶数下标的数赋给a2，len2个
    solve(len1,a1)
    solve(len2,a2)
    for(int i = 1; i <= len1; i ++) a[i] = a1[i];
    for(int i = 1; i <= len2; i ++) a[i + len1] = a2[i];
}

给出M个查询，1. 查询执行完solve的数组a中第i个数是什么，2. 查询原数组a中第i个数现在的位置

Input

多组数据

对于每组数据，第一行N,M(1 <= N <= 10^18, 1 <= M <= 10^5)

接下来M行，每行两个数x,y (1 <= x <= 2, 1 <= y <= n)

x = 1，查询执行完solve的数组a中第i个数是什么

x = 2, 查询原数组a中第i个数现在的位置

Output

每个查询输出一行，你的答案

Sample Input

4 2
1 2
2 4
6 4
1 1
1 5
1 3
2 4

Sample Output

3
4
1
6
3
6

Hint

例子N = 4,{1,2,3,4} -> {1,3} + {2,4} = {1,3,2,4}
N = 6,{1,2,3,4,5,6}-> {1,3,5} + {2,4,6} = {1,5} + {3} + {2,6} + {4} = {1,5,3,2,6,4}

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <string.h>
 5 #include <algorithm>
 6 using namespace std;
 7 #define ll long long
 8 ll fun2(ll y,ll n)
 9 {
10     if(y==1)
11     return 1;
12     if(y&1)
13     {
14         return fun2(((y>>1)+1),((n+1)>>1));
15     }
16     else
17     {
18         return ((n+1)>>1)+fun2(y>>1,n>>1);
19     }
20 }
21 ll fun1(ll y,ll n)
22 {
23     if(y==1)
24     return 1;
25     ll m=(n+1)>>1;
26     if(y<=m)
27     {
28         return (fun1(y,m)<<1)-1;
29     }
30     else
31     {
32        return  (fun1(y-m,n>>1)<<1);
33     }
34 }
35 int main()
36 {
37     ll n,x,y,ans;
38     int m,i,j;
39     while(scanf("%lld%d",&n,&m)!=EOF)
40     {
41         for(i=0;i<m;i++)
42         {
43             scanf("%lld%lld",&x,&y);
44             if(x==1)
45             {
46                printf("%lld\n",fun1(y,n));
47             }
48             else
49             {
50                 printf("%lld\n",fun2(y,n));
51             }
52         }
53     }
54     return 0;
55 }

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