LightOJ 1052

Zibon just started his courses in Computer science. After having some lectures on programming courses he fell in love with strings. He started to play with strings and experiments on them. One day he started a string of arbitrary (of course positive) length consisting of only {a, b}. He considered it as 1st string and generated subsequent strings from it by replacing all the b's with ab and all the a's with b. For example, if he ith string is abab, (i+1)th string will be b(ab)b(ab) = babbab. He found that the Nth string has length X and Mth string has length Y. He wondered what will be length of the Kth string. Can you help him?

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case begins with five integers N, X, M, Y, K. (0 < N, M, X, Y, K < 109 and N ≠ M).

Output

For each case print one line containing the case number and L which is the desired length (mod 1000000007) or the string "Impossible" if it's not possible.

Sample Input

2
3 16 5 42 6
5 1 6 10 9

Output for Sample Input

Case 1: 68
Case 2: Impossible

题目大意就是说有一个初始字符串仅包含a和b两个字符。经过Ｎ次变换字符串长度变为Ｘ，经过Ｍ次变换字符串长度变为Ｙ，现在求经过Ｋ次变换后的字符串长度。（每次变换，将字符串中的a变为b，将字符串中b变为ab）。

假设初始字符串中字符a,b的数量分别为ｘ和ｙ。

初始状态字符```a```,```b```的数量为ｘ和ｙ
经过１次变换后为　x + 2 * y
经过２次变换后为　2 * x + 3 * y
经过３次变换后为　3 * x + 5 * y
经过４次变换后为　5 * x + 8 * y

发现规律了吗？
经过几次变换后，发现变换后的系数与斐波那契数列有关。
设斐波那契数列是从1开始

1　1　2　3　5　8　13　21　...

经过Ｎ次变换后,也就是第Ｎ＋１个字符串。

g(N+1) = f(N) * x + f(N + 1) * y;

然后有这样的两个式子即可求得ｘ和ｙ，然后在求得第ｋ个字符串长度。

代码如下：

#include<bits/stdc++.h>
using namespace std;

typedef long long  ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
const ll mod = 1000000007;
bool flag;

mat mul(mat &a, mat &b)
{
    mat c(a.size(), vec(b[0].size()));
    for(int i=0; i<a.size(); ++ i)
    {
        for(int k=0; k<b.size(); ++ k)
        {
            for(int j=0; j<b[0].size(); ++ j)
            {
                c[i][j] = c[i][j] + a[i][k]*b[k][j];
                if(c[i][j] >= mod)
                {
                    flag = true;
                    c[i][j] %= mod;
                }
            }
        }
    }
    return c;
}

mat pow(mat a, ll n)
{
    mat b(a.size(), vec(a.size()));
    for(int i=0; i<a.size(); ++ i)
        b[i][i] = 1;
    while(n > 0)
    {
        if(n & 1)
            b = mul(b, a);
        a = mul(a, a);
        n >>= 1;
    }
    return b;
}

void solve(int cases)
{
    ll n, x, m, y, k;
    cin >> n >> x >> m >> y >> k;

    mat a(2, vec(2));
    a[0][0] = 1, a[0][1] = 1;
    a[1][0] = 1, a[1][1] = 0;
    flag = false;
    mat b(pow(a, n));
    mat c(mul(b, a));
    mat d(pow(a, m));
    mat e(mul(d, a));

    cout << "Case " << cases << ": ";

    if(flag || ((d[1][0] * x - b[1][0] * y) % (c[1][0] * d[1][0] - b[1][0] * e[1][0])) || ((c[1][0] * y - e[1][0] * x) % (c[1][0] * d[1][0] - e[1][0] * b[1][0])))
    {
        cout << "Impossible" << endl;
        return;
    }

    ll p = (c[1][0] * y - e[1][0] * x) / (c[1][0] * d[1][0] - e[1][0] * b[1][0]);
    ll q = (d[1][0] * x - b[1][0] * y) / (c[1][0] * d[1][0] - b[1][0] * e[1][0]);

    if(p < 0 || q < 0)
    {
        cout << "Impossible" << endl;
        return;
    }

    p %= mod;
    q %= mod;

    b = pow(a, k);
    c = mul(b, a);

    ll ans = ((p * b[1][0]) % mod + (q * c[1][0]) % mod) % mod;
    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    //freopen("out.txt", "w", stdout);
    int t;
    cin >> t;
    for(int i=1; i<=t; ++ i)
        solve(i);
    return 0;
}