time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
Examples
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.

【題目鏈接】:http://codeforces.com/contest/602/problem/A

【題解】

把它們都轉(zhuǎn)換成10進制再比較就好.
40^10不會爆LL

【完整代碼】

#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define rei(x) scanf("%d",&x) #define rel(x) scanf("%I64d",&x)typedef pair<int,int> pii; typedef pair<LL,LL> pll;const int MAXN = 10+5; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0);int n,bx,m,by; LL a[MAXN];int main() {//freopen("F:\\rush.txt","r",stdin);rei(n);rei(bx);rep2(i,n-1,0)rel(a[i]);LL x = 0;LL now = 1;rep1(i,0,n-1){x += now*a[i];now = now * bx;}rei(m);rei(by);rep2(i,m-1,0)rel(a[i]);LL y = 0;now = 1;rep1(i,0,m-1){y += now*a[i];now = now * by;}if (x==y)putchar('=');elseif (x < y)putchar('<');elseputchar('>');return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/AWCXV/p/7626809.html

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