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 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 a single character (quotes for clarity):
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.
题意:给你一个n和bx,n表示转换为bx进制后的位数,然后输入n个数,然后给你一个m和by,m表示转换为by进制后的位数,然后输入m个数,比较两个数的大小
思路:直接全部都转换为10进制,然后比较大小就行了
总结:水题,看懂样例就能写了
ac代码:
#include<stdio.h> #include<math.h> #include<string.h> #include<stack> #include<queue> #include<vector> #include<iostream> #include<algorithm> #define MAXN 200010 #define LL long long #define ll __int64 #define INF 0xfffffff #define mem(x) memset(x,0,sizeof(x)) #define PI acos(-1) using namespace std; ll fun(ll a,ll b) { ll ans=1; while(b) { if(b%2) ans=ans*a; a=a*a; b/=2; } return ans; } int main() { ll n,m,x,y,i,a; while(scanf("%I64d%I64d",&n,&x)!=EOF) { ll num1=0; for(i=n-1;i>=0;i--) { scanf("%I64d",&a); num1+=fun(x,i)*a; } scanf("%I64d%I64d",&m,&y); ll num2=0; for(i=m-1;i>=0;i--) { scanf("%I64d",&a); num2+=fun(y,i)*a; } if(num1>num2) printf(">\n"); else if(num1==num2) printf("=\n"); else printf("<\n"); } return 0; }