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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}

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 #define ll long long
 ll fun2(ll y,ll n)
 {
     if(y==)
     return ;
     if(y&)
     {
         return fun2(((y>>)+),((n+)>>));
     }
     else
     {
         return ((n+)>>)+fun2(y>>,n>>);
     }
 }
 ll fun1(ll y,ll n)
 {
     if(y==)
     return ;
     ll m=(n+)>>;
     if(y<=m)
     {
         return (fun1(y,m)<<)-;
     }
     else
     {
        return  (fun1(y-m,n>>)<<);
     }
 }
 int main()
 {
     ll n,x,y,ans;
     int m,i,j;
     while(scanf("%lld%d",&n,&m)!=EOF)
     {
         for(i=;i<m;i++)
         {
             scanf("%lld%lld",&x,&y);
             if(x==)
             {
                printf("%lld\n",fun1(y,n));
             }
             else
             {
                 printf("%lld\n",fun2(y,n));
             }
         }
     }
     return ;
 }