PROJECT EULER SOLUTION # 65

Solution to problem number 65 of Project Euler
Question # 65

The square root of 2 can be written as an infinite continued fraction.

2 = 1 +	1
	2 +	1
		2 +	1
			2 +	1
				2 + ...

The infinite continued fraction can be written, 2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, 23 = [4;(1,3,1,8)].
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for 2.

1 +	1	= 3/2
	2

1 +	1		= 7/5
	2 +	1
		2

1 +	1			= 17/12
	2 +	1
		2 +	1
			2

1 +	1				= 41/29
	2 +	1
		2 +	1
			2 +	1
				2

Hence the sequence of the first ten convergents for 2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10^th convergent is 1+4+5+7=17.
Find the sum of digits in the numerator of the 100^th convergent of the continued fraction for e.

Solution:-

/****************************************************************************/
#include<stdio.h>
#include<stdlib.h>
#define max 100
int n[max]={0},pn[max]={0};
void fun(int);
void print(int*);
void assignArray(int*,int*);
void add(int*,int*);

int main()
{
    int i,sum=0;
    for(i=1;i<=100;i++)
    {
        fun(i);
        /*
        printf("i=%d=>",i);
        print(n);
        */
    }
    for(i=0;i<max;i++)
        sum+=n[i];
    printf("answer = %d\n",sum);
    return 0;
}

void print(int *arr)
{
    int flag=0,i;
    for(i=0;i<max;i++)
    {
        if(arr[i] || flag)
        {
            flag=1;
            printf("%d",arr[i]);
        }
    }
    printf("\n");
}

void add(int*a,int*b)
{
    int i;
    for(i=max-1;i>0;i--)
    {
        a[i]+=b[i];
        if(a[i]>9)
        {
            a[i]%=10;
            a[i-1]+=1;
        }
    }
}

void assignArray(int*a,int*b)
{
    int i;
    for(i=0;i<max;i++)
        a[i]=b[i];
}

void fun(int x)
{
    long long int flag;
    int i,tmpn[max]={0},tmpArr[max]={0};
    if(x==1)
    {
        n[max-1]=2;
        return;
    }
    if(x==2)
    {
        pn[max-1]=2;
        n[max-1]=3;
        return;
    }

    if(x%3==0)
    {
        assignArray(tmpn,n);
        flag=(x/3)*2;
   
        for(i=0;i<flag;i++)
            add(tmpArr,n);

        add(tmpArr,pn);
        assignArray(n,tmpArr);

        assignArray(pn,tmpn);
        return;
    }

    assignArray(tmpn,n);
    add(n,pn);
    assignArray(pn,tmpn);
}
/****************************************************************************/