Frog and Portal HihoCoder - 1873 （二进制拆分）

A small frog wants to get to the other side of a river. The frog is initially located at one bank of the river (position 0) and wants to get to the other bank (position 200). Luckily, there are 199 leaves (from position 1 to position 199) on the river, and the frog can jump between the leaves. When at position p, the frog can jump to position p+1 or position p+2.

How many different ways can the small frog get to the bank at position 200? This is a classical problem. The solution is the 201st number of Fibonacci sequence. The Fibonacci sequence is constructed as follows: F1=F2=1;Fn=Fn-1+Fn-2.

Now you can build some portals on the leaves. For each leaf, you can choose whether to build a portal on it. And you should set a destination for each portal. When the frog gets to a leaf with a portal, it will be teleported to the corresponding destination immediately. If there is a portal at the destination, the frog will be teleported again immediately. If some portal destinations form a cycle, the frog will be permanently trapped inside. Note that You cannot build two portals on the same leaf.

Can you build the portals such that the number of different ways that the small frog gets to position 200 from position 0 is M?

Input

There are no more than 100 test cases.

Each test case consists of an integer M, indicating the number of ways that the small frog gets to position 200 from position 0. (0 ≤ M < 232)

Output

For each test case:

The first line contains a number K, indicating the number of portals.

Then K lines follow. Each line has two numbers ai and bi, indicating that you place a portal at position ai and it teleports the frog to position bi.

You should guarantee that 1 ≤ K, ai, bi ≤ 199, and ai ≠ aj if i ≠ j. If there are multiple solutions, any one of them is acceptable.

Sample Input

0
1
5

Sample Output

2
1 1
2 1
2
1 199
2 2
2
4 199
5 5

题意 ：一只青蛙想从0 到 200 ，1 到 199 都有荷叶  ，你可以放置传送门   让 A传送到B  如果 A到A 那么形成自环                

  现在给你青蛙到200应该有的方案数 ，求怎么放置传送门

解 ：二进制拆分 如果需要该位的二进制那么连到 199 如果不需要则自环

位置	1	2	3	4	5	6	7	8	9	10	11	12	13	14
方案数/传送门	自环/连199	1	1	2	自	2	自/199	2	2	4	自	4	自/199	4

 

#include <stdio.h>
#include <algorithm>
#include<string.h>
using namespace std;
#define ll long long
int main()
{
    ll n;
    while(~scanf("%lld",&n))
    {
        printf("65\n");
        if(n&(1ll<<0)) printf("1 199\n");
        else printf("1 1\n");
        for(ll i=1;i<32;i++)
        {
          printf("%lld %lld\n",5+(i-1)*6,5+(i-1)*6);
          if(n&(1ll<<i)) printf("%lld 199\n",7+(i-1)*6);
          else printf("%lld %lld\n",7+(i-1)*6,7+(i-1)*6);
        }
        printf("197 197\n");
        printf("198 198\n");
    }
}