Project Euler 6~10
Problem 6
The sum of the squares of the first ten natural numbers is,
The square of the sum of the first ten natural numbers is,
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 
385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
#include<stdio.h>
int main()
{
int i;
long long ans;
ans = 0;
for(i = 1;i <= 100;i++)
ans += (i * i);
ans = 5050 * 5050 - ans;
printf("%lld\n",ans);
return 0;
}
Problem 7
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10001st prime number?
#include<stdio.h>
#include<string.h>
#define Max 110000
int main()
{
int a[Max+1];
int temp,n,i,cnt;
memset(a,0,sizeof(a)) ;
a[0] = a[1] = 1;
for (i = 2;i <= Max;i++)
{
if (a[i] == 0)
{
temp = 2 * i;
while (temp <= Max)
{
a[temp] = 1;
temp += i;
}
}
}
cnt = 0;
for (i = 2;i <= Max;i++)
{
if(a[i] == 0)
cnt++;
if(cnt == 10001)
break;
}
printf("%d\n",i);
return 0;
}
Problem 8
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
#include<iostream>
#include<string>
using namespace std;
string s = "73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450";
int main()
{
int i,j,ans,max = -1;
for(i = 0;i <= 996;i++)
{
ans = 1;
for(j = i;j <= i + 4;j++)
ans *= (s[j] - '0');
if(ans > max)
max = ans;
}
cout << max <<endl;
return 0;
}
Problem 9
A Pythagorean triplet is a set of three natural numbers, a 
b c, for which,
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
#include<stdio.h>
#define Max 1000
int main()
{
int a,b;
for(a = 1; a <= Max;a++)
for(b = a + 1;b <= Max;b++)
if(a * a + b * b == (Max - a - b) * (Max - a - b))
goto end;
end: printf("%d\n",a * b * (Max - a - b));
return 0;
}
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
#include<stdio.h>
#include<string.h>
#define Max 2000000
int main()
{
int a[Max + 1];
long temp,n,i;
long long ans;
memset(a,0,sizeof(a)) ;
a[0] = a[1] = 1;
for (i = 2;i <= Max;i++)
{
if (a[i] == 0)
{
temp = 2 * i;
while (temp <= Max)
{
a[temp] = 1;
temp += i;
}
}
}
ans = 0;
for (i = 2;i < Max;i++)
if(a[i] == 0)
ans += i;
printf("%lld\n",ans);
return 0;
}
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