Project Euler -- 48 (Self powers)

Problem : The series, 1¹ + 2² + 3³ + ... + 10¹⁰ = 10405071317.

Find the last ten digits of the series, 1¹ + 2² + 3³ + ... + 1000¹⁰⁰⁰.

My solution approach is,

first made a smaller version of the problem then I tried to solve it after solving this I was going to solve the bigger version.

// a. find out p=2^15

// b. print the last digit of p

	#include<stdio.h>
	#include<math.h>
	// a. find out p=2^15
	// b. print the last digit of p
	int main()
	{
	int ans,i,n,j;
	n=10;
	ans=2;
	for(i=2;i<=15;i++){
	ans=ans*2;
	ans=ans%10; // if I want to know the last one digit of the number of the sum of the sequence
	}
	printf("ans :%d",ans);
	return 0;
	}

view raw gistfile1.c hosted with ❤ by GitHub

	/// Euler problem 48
	#include<stdio.h>
	#include<math.h>
	int long long power(int long long n);
	int main()
	{
	int long long ans,i,j;
	ans=0;
	for(i=1;i<=1000;i++){
	ans= ans+power(i);
	ans=ans%10000000000;
	}
	printf("ans :%lld",ans);
	return 0;
	}
	int long long power(int long long n)
	{
	int i;
	int long long p=1;
	for(i=1;i<=n;i++){
	p=p*n;
	p=p%10000000000;
	}
	return p;
	}

view raw euler 48 solution hosted with ❤ by GitHub