Project Euler 6~10

whitedeath posted @ 2010年5月18日 07:17 in Project Euler with tags 数论 ACM , 1836 阅读

Problem 6

14 December 2001
 

The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025

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.

Answer:
25164150
#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

28 December 2001

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?

Answer:
104743

 

#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

11 January 2002


 

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
 


 

Answer:
40824
 
#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

25 January 2002
 

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Answer:
31875000
#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

08 February 2002
 

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.
 

Answer:
142913828922
#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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