【CodeForces 602A】C - 特别水的题3-Two Bases

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/C

Description

After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers X and Y realised that they have different bases, which complicated their relations.

You're given a number X represented in base b_x and a number Y represented in base b_y. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and b_x (1 ≤ n ≤ 10, 2 ≤ b_x ≤ 40), where n is the number of digits in the b_x-based representation of X.

The second line contains n space-separated integers x₁, x₂, ..., x_n (0 ≤ x_i < b_x) — the digits of X. They are given in the order from the most significant digit to the least significant one.

The following two lines describe Y in the same way: the third line contains two space-separated integers m and b_y (1 ≤ m ≤ 10, 2 ≤ b_y ≤ 40, b_x ≠ b_y), where m is the number of digits in the b_y-based representation of Y, and the fourth line contains m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y) — the digits of Y.

There will be no leading zeroes. Both X and Y will be positive. All digits of both numbers are given in the standard decimal numeral system.

Output

Output a single character (quotes for clarity):

• '<' if X < Y
• '>' if X > Y
• '=' if X = Y

Sample Input

Input

6 2
1 0 1 1 1 1
2 10
4 7

Output

=

Input

3 3
1 0 2
2 5
2 4

Output

<

Input

7 16
15 15 4 0 0 7 10
7 9
4 8 0 3 1 5 0

Output

>

Hint

In the first sample, X = 101111₂ = 47₁₀ = Y.

In the second sample, X = 102₃ = 21₅ and Y = 24₅ = 112₃, thus X < Y.

In the third sample,   and  Y = 4803150₉ . We may notice that  X  starts with much larger digits and  b_x  is much larger than  b_y , so  X  is clearly larger than  Y . 

 

转换进制，一边读入一边转为十进制，再比较大小。

#include
long long x,y,bx,by,num,decx,decy;
int main(){
    scanf("%lld%lld",&x,&bx);
    for(int i=0;i1;i++){
        scanf("%lld",&num);
        decx=(decx+num)*bx;
    }
    scanf("%lld",&num);
    decx=decx+num;
    scanf("%lld%lld",&y,&by);
    for(int i=0;i1;i++){
        scanf("%lld",&num);
        decy=(decy+num)*by;
    }
    scanf("%lld",&num);
    decy=decy+num;

    if(decx>decy)printf(">\n");
    else if(decx"<\n");
    else printf("=\n");
    return 0;
}

 

　　

 

转载于:https://www.cnblogs.com/flipped/p/5045267.html