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 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 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.
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 }