CodeForces 197B Limit

Limit

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

Submit Status Practice CodeForces 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 and m (0 ≤ n, m ≤ 100) — degrees of polynomials P(x) and Q(x)correspondingly.

The second line contains n + 1 space-separated integers — the factors of polynomial P(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 polynomial Q(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 1
1 1 1
2 5

Output

Infinity

Input

1 0
-1 3
2

Output

-Infinity

Input

0 1
1
1 0

Output

0/1

Input

2 2
2 1 6
4 5 -7

Output

1/2

Input

1 1
9 0
-5 2

Output

-9/5

Hint

题意：求极限。

题解：首先

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.

的极限P(x)=a0*x^n，Q(x)=b0*x^m,x->INF.

然后直接比較即可了。

注：gcd(-2,4)=-2.

#include<cstdio>
#include<iostream>
#include<cstring>

using namespace std;
#define N 110

int n,m,a[N],b[N];

int gcd(int a,int b) {
    return b==0?a:gcd(b,a%b);
}

int main() {
    //printf("%d\n",gcd(-2,4));
    while(cin>>n>>m) {
        n++,m++;
        for(int i=0; i<n; i++) {
            scanf("%d",&a[i]);
        }
        for(int j=0; j<m; j++) {
            scanf("%d",&b[j]);
        }
        if(n==m) {
            if(b[0]<0) {
                b[0]=-b[0];
                a[0]=-a[0];
            }
            int x=gcd(a[0],b[0]);
            if(x<0)x=-x;
            printf("%d/%d\n",a[0]/x,b[0]/x);
        } else if(n>m) {
            if(a[0]>0&&b[0]>0||(a[0]<0&&b[0]<0))printf("Infinity\n");
            else       printf("-Infinity\n");
        } else {
            printf("0/1\n");
        }
    }
}