poj 2116 Death to Binary? 模拟

Death to Binary?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1707		Accepted: 529

Description

The group of Absurd Calculation Maniacs has discovered a great new way how to count. Instead of using the ordinary decadic numbers, they use Fibonacci base numbers. Numbers in this base are expressed as sequences of zeros and ones similarly to the binary numbers, but the weights of bits (fits?) in the representation are not powers of two, but the elements of the Fibonacci progression (1, 2, 3, 5, 8,... - the progression is defined by F0 = 1, F1 = 2 and the recursive relation Fn = Fn-1 + Fn-2 for n >= 2).

For example 1101001Fib = F0 + F3 + F5 + F6 = 1 + 5 + 13 + 21 = 40.

You may observe that every integer can be expressed in this base,
but not necessarily in a unique way - for example 40 can be also
expressed as 10001001Fib. However, for any integer there is a
unique representation that does not contain two adjacent digits 1 - we
call this representation canonical. For example 10001001Fib is a canonical Fibonacci representation of 40.

To prove that this representation of numbers is superior to the
others, ACM have decided to create a computer that will compute in
Fibonacci base. Your task is to create a program that takes two numbers
in Fibonacci base (not necessarily in the canonical representation) and
adds them together.

Input

The
input consists of several instances, each of them consisting of a single
line. Each line of the input contains two numbers X and Y in Fibonacci
base separated by a single space. Each of the numbers has at most 40
digits. The end of input is not marked in any special way.
Output

The output for each instance should be formated as follows:

The first line contains the number X in the canonical
representation, possibly padded from left by spaces. The second line
starts with a plus sign followed by the number Y in the canonical
representation, possibly padded from left by spaces. The third line
starts by two spaces followed by a string of minus signs of the same
length as the result of the addition. The fourth line starts by two
spaces immediately followed by the canonical representation of X + Y.
Both X and Y are padded from left by spaces so that the least
significant digits of X, Y and X + Y are in the same column of the
output. The output for each instance is followed by an empty line.

Sample Input

11101 1101
1 1

Sample Output

100101
+   10001
-------
1001000

1
+  1
--
10

Source

CTU Open 2004
题意：给你一个两个字符串，一个字符串的值等于为1位置的斐波那契的和，比如1101001Fib = F0 + F3 + F5 + F6 = 1 + 5 + 13 + 21 = 40，一个值可能有多种不同的写法，需要改成没有相邻的1的写法， 写成加法的式子；
思路：模拟，坑点 0 0；和前导0；

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define ll long long
#define esp 1e-13
const int N=1e4+10,M=1e6+50000,inf=1e9+10,mod=1000000007;
string s1,s2,s3;
ll a
;
void init()
{
a[0]=1;
a[1]=2;
for(int i=2;i<=50;i++)
a[i]=a[i-1]+a[i-2];
}
ll getnum(string aa)
{
int x=aa.size();
ll sum=0;
for(int i=0;i<x;i++)
if(aa[i]=='1')
sum+=a[i];
return sum;
}
void check(ll x,string &str)
{
int i;
for(i=50;i>=0;i--)
if(x>=a[i])
break;
for(int t=i;t>=0;t--)
if(x>=a[t])
{
str+='1';
x-=a[t];
}
else
str+='0';
if(i<0)
str+='0';
}
int main()
{
int x,y,i,z,t;
init();
while(cin>>s1>>s2)
{

reverse(s1.begin(),s1.end());
reverse(s2.begin(),s2.end());
ll num1=getnum(s1);
ll num2=getnum(s2);
ll num3=num1+num2;
s1.clear();
s2.clear();
s3.clear();
check(num1,s1);
check(num2,s2);
check(num3,s3);
printf("  ");for(i=0;i<s3.size()-s1.size();i++)printf(" ");cout<<s1<<endl;
printf("+ ");for(i=0;i<s3.size()-s2.size();i++)printf(" ");cout<<s2<<endl;
printf("  ");for(i=0;i<s3.size();i++)printf("-");cout<<endl;
printf("  ");cout<<s3<<endl;
cout<<endl;
}
return 0;
}