A. Two Bases
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 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
>
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, 
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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
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<n1; i++)
cin>>a[i];
LL sum1 = 0, sum2 = 0;
for(int i=n1-1; i>=0; i--)
{
LL sum = 1;
for(int j=0; j<n1-1-i; j++)
sum *= num1;
sum1 += a[i]*sum;
}
///cout<<sum1<<endl;
cin>>n2>>num2;
for(int i=0; i<n2; i++)
cin>>b[i];
for(int i=n2-1; i>=0; i--)
{
LL sum = 1;
for(int j=0; j<n2-1-i; j++)
sum *= num2;
sum2 += b[i]*sum;
}
///cout<<sum2<<endl;
if(sum1 == sum2)
puts("=");
else if(sum1 < sum2)
puts("<");
else
puts(">");
return 0;
}
/**
10 16
15 15 4 0 0 0 0 7 10 9
7 9
4 8 0 3 1 5 0
*/