時間限制:1000ms

單點時限:1000ms

記憶體限制:512MB

描述

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?

輸入

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)

輸出

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.

樣例輸入

0
1
5

樣例輸出

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

有一隻青蛙，從０的位置跳到２００，每次跳躍的距離是１或２。現在你可以在某些葉子上設定傳送點，將其直接傳送到某個點，最後使得青蛙到２００的方案數位M，輸出你設計的方案。

前面很容易控制成為只有一種方案的方法，例如在奇數點設定傳送位置，那麼這樣前面的點只有一種走法，然後就是控制後米娜的方案數和是M就ｏｋ啦，需要注意的是任意一個數，總能由幾個不同位置的斐波那契陣列成，需要從大到小遍歷減去就可以。這樣前面的奇數點傳送２００－ｆｉｂ，最後一個奇數點後面的偶數直接自環，這樣前面只有一條路，後面相加等於M輸出最後的答案和跟方案。

程式碼實現：

/*
Look at the star
Look at the shine for U
*/
#include<bits/stdc++.h>
#define ll long long
#define PII pair<int,int>
#define sl(x) scanf("%lld",&x)
using namespace std;
const int N = 1e6+5;
const ll mod = 1e18;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
ll fib[N];
int main()
{
    ll n,i,j,t,cnt;
    fib[0] = 1;
    fib[1] = 1;
    ll maxx = fpow(2,32);
    for(cnt = 2;;cnt++)
    {
        fib[cnt] = fib[cnt-1]+fib[cnt-2];
        if(fib[cnt] > maxx) break;
    }
    while(~sl(n))
    {
        if(n == 0)
        {
            printf("2\n1 1\n2 1\n");
            continue;
        }
        ll sum = 1;
        ll pos = upper_bound(fib,fib+cnt,n)-fib;
        pos--;
        vector <ll> vec;
        for(i = pos;i >= 0;i--)
        {
            if(n >= fib[i])
            {
                vec.push_back(i);
                sum++;
                n -= fib[i];
            }
        }
        printf("%lld\n",sum);
        ll tt = 1,ans = 0;
        for(i = 0;i < vec.size();i++)
        {
            printf("%lld %lld\n",tt,200-vec[i]);
            ans += fib[vec[i]];
            tt += 2;
        }
        printf("%lld %lld\n",tt-1,tt-1);
    }
    return 0;
}