zhber
有好多做过的题没写下来,如果我还能记得就补吧
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 thebx-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.

 

闲来无事自己找了一场div2做做

第一题进制转换没开longlong就跪了一次。。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define pi 3.1415926535897932384626433832795028841971
16 using namespace std;
17 inline LL read()
18 {
19     LL x=0,f=1;char ch=getchar();
20     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
21     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
22     return x*f;
23 }
24 inline void write(LL a)
25 {
26     if (a<0){printf("-");a=-a;}
27     if (a>=10)write(a/10);
28     putchar(a%10+'0');
29 }
30 inline void writeln(LL a){write(a);printf("\n");}
31 LL s[110];
32 LL a,b; 
33 inline void getn(LL &a)
34 {
35     LL x=read(),y=read(),z=1ll;
36     for (int i=1;i<=x;i++)s[i]=read();
37     for (int i=x;i>=1;i--)a+=s[i]*z,z*=y;
38 }
39 int main()
40 {
41     getn(a);
42     getn(b);
43     if (a==b)puts("=");
44     if (a<b)puts("<");
45     if (a>b)puts(">");
46 }
cf602A

 

posted on 2016-08-07 12:19  zhber  阅读(295)  评论(0编辑  收藏  举报