【42.59%】【codeforces 602A】Two Bases
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time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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
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.
【题目链接】:http://codeforces.com/contest/602/problem/A
【题解】
把它们都转换成10进制再比较就好.
40^10不会爆LL
【完整代码】
#include <bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%I64d",&x)typedef pair<int,int> pii;typedef pair<LL,LL> pll;const int MAXN = 10+5;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);int n,bx,m,by;LL a[MAXN];int main(){ //freopen("F:\\rush.txt","r",stdin); rei(n);rei(bx); rep2(i,n-1,0) rel(a[i]); LL x = 0; LL now = 1; rep1(i,0,n-1) { x += now*a[i]; now = now * bx; } rei(m);rei(by); rep2(i,m-1,0) rel(a[i]); LL y = 0; now = 1; rep1(i,0,m-1) { y += now*a[i]; now = now * by; } if (x==y) putchar('='); else if (x < y) putchar('<'); else putchar('>'); return 0;}
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