A. Two Basestime 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.
此题坑点“pow函数会损失精度！
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;typedef long long int LL;
int a[50], b[50];LL pow1(int a, int b) {
LL res = 1;
while (b --)
res *= a;
return res;
}int main() {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
int n, m, k1, k2;
cin >> n >> k1;
for (int i = n-1; i >=0; i--)
cin >> a[i];
LL res1 = 0;
for (int i = 0; i < n; i++)
res1 += a[i] * pow1(k1, i);
cin >> m >> k2;
for (int i = m - 1; i >= 0; i--)
cin >> b[i];
LL res2 = 0;
for (int i = 0; i < m; i++)
res2 += b[i] * pow1(k2, i);
//cout << res1 << endl << res2 << endl;if (res1 == res2)
cout << "=" << endl;
else if (res1 > res2) cout << ">" << endl;
else cout << "<" << endl;
return 0;
}
