HihoCoder - 1873 2018ICPC北京站 D. Frog and Portal(构造)

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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?

输入

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)

输出

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.

样例输入

0
1
5

样例输出

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

思路

先说题意，给你一个坐标轴，有一个青蛙，要从 $0$这个位置跳到 $200$这个位置，他的起始位置为 $0$.

青蛙每次可以向后跳 $1$步或者 $2$步,现在你可以给某个点放置一个传送门，规定传送门的起始和终止位置，那么青蛙到达这个点后会从传送门的起点瞬移到传送门的终点，规定一个点最多只能放置一个传送门，如果一个传送门的终点有一个传送门，那么青蛙可以连续传送。

现在给出你一个方案数 $M$，你需要在这个坐标轴上放置传送门，使得青蛙从 $0$到 $200$这个位置的方案数恰好为 $M$

构造题，假设当前青蛙所在位置为 $x$，因为他每次可以向后跳1或2步，那么构造两个门(x,x+2),(x+1,x)，这样无论青蛙怎么跳，最后都会停在x+2这个位置上。那么方案数就变成了原来的一半。

所以我们可以根据 $M$的奇偶性，当 $M$为偶数的时候就构造(x,x+2),(x+1,x)使得方案数变成原来的一半，当 $M$为奇数的时候，就可以构造从当前点直接连到199，(x,199),使得方案数减一，那么方案数就变成了偶数，按照原来的方法继续构造即可，一直到方案数为1时，直接给当前位置连接到199即可。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pir;
int main()
{
    ll n;
    while (~scanf("%lld", &n))
    {
        if (n == 0)
            printf("2\n1 1\n2 1\n");
        else if (n == 1)
            printf("2\n1 199\n2 2\n");
        else
        {
            vector<pir> v;
            ll now = 1;
            while (n != 1)
            {
                if (n & 1)
                {
                    v.push_back(pir(now, 199));
                    now += 2;
                    v.push_back(pir(now, now + 2));
                    v.push_back(pir(now + 1, now));
                    now += 3;
                }
                else
                {
                    v.push_back(pir(now, now + 2));
                    v.push_back(pir(now + 1, now));
                    now += 3;
                }
                n >>= 1;
            }
            v.push_back(pir(now - 1, 199));
            printf("%lld\n", v.size());
            for (auto ans : v)
                printf("%lld %lld\n", ans.first, ans.second);
        }
    }
    return 0;
}