HDU 4814 golden radio base

Question:

Convert a 10-digit number to X-digit number x = (1 + √ 5)/2

Ideas:

The game was scared by the logic of the number. Actually, it's not necessary because n = N * 1 = N * x ^ 0.

Since the zero power of X is equal to 1, we can regard N as a single digit of the corresponding X-base number. Now we need to convert this number into binary.

The formula given by the question can multiply the power of X at the same time on both sides. Then the two formulas become

X ^ (I + 1) + x ^ I = x ^ (I + 2)

2 * x ^ I = x ^ (I + 1) + x ^ (I-2)

Use these two formulas to constantly change the number until the number does not change.

Code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 100int a[N*2];int n,u,v;int main(){    int i,k,flag;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        a[N]=n;        do        {            flag=0;            for(i=0;i<N*2-2;i++)            {                if(a[i]&&a[i+1])                {                    k=min(a[i],a[i+1]);                    a[i]-=k;                    a[i+1]-=k;                    a[i+2]+=k;                    flag=1;                }            }            for(i=2;i<N*2-1;i++)            {                if(a[i]>1)                {                    k=a[i]/2;                    a[i]%=2;                    a[i-2]+=k;                    a[i+1]+=k;                    flag=1;                }            }        }while(flag);        for(u=2*N-1;u>N&&!a[u];u--);        for(v=0;v<N&&!a[v];v++);        for(i=u;i>=N;i--) printf("%d",a[i]);        if(v!=N)        {            printf(".");            for(i=N-1;i>=v;i--) printf("%d",a[i]);        }        printf("\n");    }    return 0;}