Codeforces Round #333 (Div. 2) A. Two Bases-白红宇

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 bx and a number Y represented in base by. Compare those two numbers.

Input

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

The second line contains n space-separated integers x1, x2, ..., xn (0 ≤ xi < bx) — 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 by (1 ≤ m ≤ 10,2 ≤ by ≤ 40, bx ≠ by), where m is the number of digits in the by-based representation of Y, and the fourth line contains m space-separated integers y1, y2, ..., ym (0 ≤ yi < by) — 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 test(s) 
        input 
6 21 0 1 1 1 12 104 7
        output 
=
        input 
3 31 0 22 52 4
        output 
<
        input 
7 1615 15 4 0 0 7 107 94 8 0 3 1 5 0
        output 
>

Note

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, $\text{[math]}$ and Y = 48031509. We may notice that X starts with much larger digits and bx is much larger than by, so X is clearly larger than Y.

题意：

给你两个数，让你比较大小，但是进制不同，位数也不一定相同

解题思路：

直接模拟做，全都化成十进制数在做。。

上代码

#include 
     
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       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 1e5+5;const int mod = 1e9+7;const double eps = 1e-10;const int INF = 0x3f3f3f3f;LL gcd(LL a, LL b){ if(b == 0) return a; return gcd(b, a%b);}LL a[maxn], b[maxn];int main(){ LL n1, n2, num1, num2; cin>>n1>>num1; for(int i=0; i
              
               >a[i]; LL sum1 = 0, sum2 = 0; for(int i=n1-1; i>=0; i--) { LL sum = 1; for(int j=0; j
               
                >n2>>num2; for(int i=0; i
                
                 >b[i]; for(int i=n2-1; i>=0; i--) { LL sum = 1; for(int j=0; j
                 
                  <"); else puts(">"); return 0;}/**10 1615 15 4 0 0 0 0 7 10 97 94 8 0 3 1 5 0*/