codeforces Limit

C - Limit

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeCodeForces 197B

Description

You are given two polynomials:

• P(x) = a₀·xⁿ + a₁·x^n - 1 + ... + a_n - 1·x + a_n and
• Q(x) = b₀·x^m + b₁·x^m - 1 + ... + b_m - 1·x + b_m.

Calculate limit .

Input

The first line contains two space-separated integers n andm (0 ≤ n, m ≤ 100) — degrees of polynomialsP(x) and Q(x) correspondingly.

The second line contains n + 1 space-separated integers — the factors of polynomialP(x): a₀,a₁, ..., a_n - 1, a_n( - 100 ≤ a_i ≤ 100, a₀ ≠ 0).

The third line contains m + 1 space-separated integers — the factors of polynomialQ(x): b₀,b₁, ..., b_m - 1, b_m( - 100 ≤ b_i ≤ 100, b₀ ≠ 0).

Output

If the limit equals  + ∞, print "Infinity" (without quotes). If the limit equals - ∞, print "-Infinity" (without the quotes).

If the value of the limit equals zero, print "0/1" (without the quotes).

Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where p is the — numerator, q(q > 0) is the denominator of the fraction.

Sample Input

Input

2 11 1 12 5

Output

Infinity

Input

1 0-1 32

Output

-Infinity

Input

0 111 0

Output

0/1

Input

2 22 1 64 5 -7

Output

1/2

Input

1 19 0-5 2

Output

-9/5

Hint

Let's consider all samples:

1. 
2. 
3. 
4. 
5. 

You can learn more about the definition and properties of limits if you follow the link:http://en.wikipedia.org/wiki/Limit_of_a_function

思路：

求 n趋于无穷大的时候函数的极限

1.长度n < m时  全部为0

2.n > m 时  分符号相反和相同两种情况

3 n = m 时 ,需要求出通分之后的结果，所以都除以最大公约数，注意同为负时，不输出负号

/*********************************************** * Author: fisty * Created Time: 2015/2/4 20:26:30 * File Name   : 3_C.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define MAX_N 110#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;#define FOR(i, a, b) for(int i = a;i < b; i++)int P[MAX_N];int Q[MAX_N];int gcd(int a, int b){    return b == 0 ? a : gcd(b , a % b);}int main() {    //freopen("in.cpp", "r", stdin);    cin.tie(0);    ios::sync_with_stdio(false);    int n, m;       cin >> n >> m;    for(int i = 0;i <= n; i++){        cin >> P[i];    }    for(int i = 0;i <= m; i++){        cin >> Q[i];    }    if(n > m){        if(P[0] < 0 && Q[0] < 0){            cout << "Infinity" << endl;        }else if(P[0] < 0 && Q[0] > 0){            cout << "-Infinity" << endl;        }else if(P[0] > 0 && Q[0] < 0){            cout << "-Infinity" << endl;        }else{            cout << "Infinity" << endl;        }    }else if(n < m){        cout << 0 << "/" << 1 << endl;    }else{        if(P[0] * Q[0] < 0) cout << "-";        if(P[0] < 0) P[0] *= -1;        if(Q[0] < 0) Q[0] *= -1;        int g = gcd(P[0], Q[0]);        cout << P[0] / g << "/" << Q[0] / g << endl;    }    return 0;}