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  • cf602A Two Bases

    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("
    ");}
    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
    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/5745918.html
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