CodeForces 602 A. Two Bases

题目描述：

水题，只是注意精度，还有如果你要用pow()函数来求解幂，那么注意pow()返回值为double，如果你用long long int 那么就wrong了，及时类型转换可是在大数据时还是出现数据错乱，不知道为什么。

思路：都转换成十进制然后比较就行

提后分析：

用c做十进制的输入输出格式

int %d

long int %ld

long long int %I64d

unsigned int %d

unsigned long int %ld

unsigned long long int %I64d

double 输入%lf

float 输入 %f

输出 %lf或%f都行

long double 就不知道用什么了 还是用cin吧！！！

题目：

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.

代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string>#include <stack>#include <queue>#include <algorithm>#include <math.h>
using namespace std;

long double x, ax, y, by;
long double DX, DY;//存储转换后十进制的数据

long double toDecimal( long double a, long double ax)
{
long double temp = 0;
long double d = 0;
for( int i=(int)a; i>0; i--)
{
cin>> temp;
d += temp * pow(ax, i-1);
}
return d;
}

int main()
{
while( cin>> x >> ax )
{
long double x1 = toDecimal(x, ax);
cin>> y >> by;
long double y1 = toDecimal(y, by);
if(x1 > y1)
cout<<">"<<endl;
else
if(x1 == y1)
cout<<"="<<endl;
else
cout<<"<"<<endl;
}
return 0;
}

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