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):
- '<' 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.
题意:
给你两个数,让你比较大小,但是进制不同,位数也不一定相同
解题思路:
直接模拟做,全都化成十进制数在做。。
上代码
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <set> using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long LL; typedef unsigned long long ULL; const int maxn = 1e5+5; const int mod = 1e9+7; const double eps = 1e-10; const int INF = 0x3f3f3f3f; LL gcd(LL a, LL b) { if(b == 0) return a; return gcd(b, a%b); } LL a[maxn], b[maxn]; int main() { LL n1, n2, num1, num2; cin>>n1>>num1; for(int i=0; i<n1; i++) cin>>a[i]; LL sum1 = 0, sum2 = 0; for(int i=n1-1; i>=0; i--) { LL sum = 1; for(int j=0; j<n1-1-i; j++) sum *= num1; sum1 += a[i]*sum; } ///cout<<sum1<<endl; cin>>n2>>num2; for(int i=0; i<n2; i++) cin>>b[i]; for(int i=n2-1; i>=0; i--) { LL sum = 1; for(int j=0; j<n2-1-i; j++) sum *= num2; sum2 += b[i]*sum; } ///cout<<sum2<<endl; if(sum1 == sum2) puts("="); else if(sum1 < sum2) puts("<"); else puts(">"); return 0; } /** 10 16 15 15 4 0 0 0 0 7 10 9 7 9 4 8 0 3 1 5 0 */