CF memsql Start[c]UP 2.0 A

CF memsql Start[c]UP 2.0 A

A. Golden System

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Piegirl got bored with binary, decimal and other integer based counting systems. Recently she discovered some interesting properties about number

, in particular that q2 = q + 1, and she thinks it would make a good base for her new unique system. She called it "golden system". In golden system the number is a non-empty string containing 0's and 1's as digits. The decimal value of expressiona0a1...an equals to

.

Soon Piegirl found out that this system doesn't have same properties that integer base systems do and some operations can not be performed on it. She wasn't able to come up with a fast way of comparing two numbers. She is asking for your help.

Given two numbers written in golden system notation, determine which of them has larger decimal value.

Input

Input consists of two lines — one for each number. Each line contains non-empty string consisting of '0' and '1' characters. The length of each string does not exceed 100000.

Output

Print ">" if the first number is larger, "<" if it is smaller and "=" if they are equal.

Sample test(s)

input

1000 
111 

output

< 

input

00100 
11 

output

= 

input

110 
101 

output

> 

Note

In the first example first number equals to

, while second number is approximately1.6180339882 + 1.618033988 + 1 ≈ 5.236, which is clearly a bigger number.

In the second example numbers are equal. Each of them is  ≈ 2.618.

 

思路：最开始想推每一项的公式，不行，系数太大！后来想把前面的1全部转化为后面的1，发现这样的话也会2^100000太大！

后来发现如果一个字符串中出现的第一个1比另一个字符串中的第一个1高两位的话，就是这个串大，否则转化为后面的1（也就是第i位的1等于第i-1位的1和第i-2位的1）然后再逐位判断对多10^5。

处理时要把字符串反转，放在vector里面然后reverse(),竟然超时，换做直接字符串反转函数，AC 30ms！

  1 #include<cstdio>
  2 
  3 #include<iostream>
  4 
  5 #include<cmath>
  6 
  7 #include<stdlib.h>
  8 
  9 #include<vector>
 10 
 11 #include<cstring>
 12 
 13 #include<map>
 14 
 15 #include<algorithm>
 16 
 17 #include<string.h>
 18 
 19 #define M(a,b) memset(a,b,sizeof(a))
 20 
 21 #define INF 0x3f3f3f3f
 22 
 23 
 24 
 25 using namespace std;
 26 
 27 
 28 
 29 char a[100005],b[100005];
 30 
 31 int num[100005];
 32 
 33 vector<char> v1,v2;
 34 
 35 void strRev(char *s)
 36 
 37 {
 38 
 39     char temp, *end = s + strlen(s) - 1;
 40 
 41     while( end > s)
 42 
 43     {
 44 
 45         temp = *s;
 46 
 47         *s = *end;
 48 
 49         *end = temp;
 50 
 51         --end;
 52 
 53         ++s;
 54 
 55     }
 56 
 57 }
 58 
 59 
 60 
 61 int main()
 62 
 63 {
 64 
 65     while(scanf("%s",a)==1)
 66 
 67     {
 68 
 69         scanf("%s",b);
 70 
 71         int n = max(strlen(a),strlen(b));
 72 
 73         strRev(a);
 74 
 75         strRev(b);
 76 
 77         int strla = strlen(a);
 78 
 79         int strlb = strlen(b);
 80 
 81         int tem = strlen(a)-strlen(b);
 82 
 83         if(tem < 0)
 84 
 85         {
 86 
 87             for(int i = 0;i<-tem;i++)
 88 
 89                 a[strla+i] = '0';
 90 
 91         }
 92 
 93         else
 94 
 95         {
 96 
 97             for(int i = 0;i<tem;i++)
 98 
 99                 b[strlb+i] = '0';
100 
101         }
102 
103         /*for(int i = 0;i<n;i++) cout<<a[i];
104 
105         cout<<endl;
106 
107          for(int i = 0;i<n;i++) cout<<b[i];
108 
109         cout<<endl;*/
110 
111         for(int i = 0;i<n;i++)
112 
113         {
114 
115             if(a[i] == '1' && b[i] == '1') num[i+2] = 0;
116 
117             else if(a[i] == '0' && b[i] == '0') num[i+2] = 0;
118 
119             else if(a[i] == '1' && b[i] == '0') num[i+2] = 1;
120 
121             else if(a[i] == '0' && b[i] == '1') num[i+2] = -1;
122 
123         }
124 
125         num[0] = 0;
126 
127         num[1] = 0;
128 
129         //for(int i = 0;i<n+2;i++) cout<<num[i];
130 
131         //cout<<endl;
132 
133         for(int i = n+1;i>=0;i--)
134 
135         {
136 
137             if(i==1&&num[i]==0) puts("=");
138 
139             if(num[i]==2){puts(">"); break;}
140 
141             if(num[i]==-2){puts("<"); break;}
142 
143             if(num[i] == 0) continue;
144 
145             if(num[i] == 1)
146 
147             {
148 
149                 if(num[i-1]==0||num[i-1]==1) {puts(">"); break;}
150 
151                 if(num[i-1]==-1) num[i-1]+=1, num[i-2]+=1;
152 
153             }
154 
155             if(num[i] == -1)
156 
157             {
158 
159                 if(num[i-1]==0||num[i-1]==-1) {puts("<"); break;}
160 
161                 if(num[i-1]==1) num[i-1]-=1, num[i-2]-=1;
162 
163             }
164 
165         }
166 
167     }
168 
169     return 0;
170 
171 }

 

posted @ 2014-10-19 19:49  haohaooo  阅读(200)  评论(0)  编辑  收藏  举报