A. Two Bases
1 second
256 megabytes
standard input
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.
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 a single character (quotes for clarity):
- '<' if X < Y
- '>' if X > Y
- '=' if X = Y
- 6 2
- 1 0 1 1 1 1
- 2 10
- 4 7
- =
- 3 3
- 1 0 2
- 2 5
- 2 4
- <
- 7 16
- 15 15 4 0 0 7 10
- 7 9
- 4 8 0 3 1 5 0
- >
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.
- #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;
- }
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