欧拉计划6-10题

6、Sum square difference

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.

前十个自然数的平方和是：

12 + 22 + ... + 102 = 385

前十个自然数的和的平方是：

(1 + 2 + ... + 10)2 = 552 = 3025

所以平方和与和的平方的差是3025 − 385 = 2640.

找出前一百个自然数的平方和与和平方的差。

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
  
#define N 100
  
int powplus(int n, int k)
{
    int s=1;
    while(k--)
    {
       s*=n;
    }
  return s;
}
  
int sum1(int n)
{
   return  powplus((n+1)*n/2,2);
} 
  
int sum2(int n)
{
   return (n*(n+1)*(2*n+1))/6;
}
  
void solve()
{
     printf("%d\n",sum1(N));
     printf("%d\n",sum2(N));
     printf("%d\n",sum1(N)-sum2(N));
} 
  
int main()
{
  solve();
  return 0;
}
View Code
Answer:25164150
Completed on Tue, 2 Apr 2013, 06:57
 

 7、10001st prime

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 10 001st prime number?

前六个质数是2,3,5,7,11和13，其中第6个是13.

第10001个质数是多少？

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
  
int prim(int n)
{
   int i;
   for(i=2; i*i<=n; i++)
   {
      if(n%i==0)
        return 0;
   }
   return 1;
}
  
void solve(int n)
{
  int i=2;
  int count=0;
  while(1)
  {
     if(prim(i))
     { 
       count++;
       if(count==n)
         break;
     }
     i++;
  }
  printf("%d\n",i);
}
  
  
int main()
{
  int n=10001;
  solve(n);
  return 0;
}
View Code
Answer:104743

Completed on Thu, 4 Apr 2013, 17:34

 

 8、Largest product in a series

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

找出以下这个1000位的整数中连续5个数字的最大乘积。（例如前五个数字的乘积是7*3*1*6*7=882)

#include <stdio.h>
#include <stdlib.h>

int main()
{
FILE *fp;
char *buffer;
int i=0,j=0;

fp=fopen("E://file.txt","r");

char c;

while((c=fgetc(fp))){
  if(c==EOF)
    break;
  else if(c!='\n')
    i++;
}
buffer=(char *)malloc(i*sizeof(char));
rewind(fp);
while((c=fgetc(fp))){
  if(c==EOF)
    break;
  else if(c!='\n')
    {
        *(buffer+j)=c;
        j++;
    }
}
findmax(buffer,i);
}


int findmax(char *buffer,int i)
{
    int j=0,max=0;

    for(j=0;j<i-4;j++)
    {
        if(max<((buffer[j]-'0')*(buffer[j+1]-'0')*(buffer[j+2]-'0')*(buffer[j+3]-'0')*(buffer[j+4]-'0')))
            max=((buffer[j]-'0')*(buffer[j+1]-'0')*(buffer[j+2]-'0')*(buffer[j+3]-'0')*(buffer[j+4]-'0'));

    }
printf("\n%d",max);
return 0;

}
View Code
Answer:40824
Completed on Sun, 17 Nov 2013, 11:16
 

 9、Special Pythagorean triplet

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.

一个毕达哥拉斯三元组是一个包含三个自然数的集合，a<b<c，满足条件：

a2 + b2 = c2

例如：32 + 42 = 9 + 16 = 25 = 52.

已知存在并且只存在一个毕达哥拉斯三元组满足条件a + b + c = 1000。

找出该三元组中abc的乘积。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
#include<stdbool.h>

void show()
{
    int a,b,c;
    for(a=1; a<333; a++)
    {
        for(c=333; c<500; c++)
        {
            b=1000-a-c;
            if(a*a+b*b==c*c)
            {
                printf("%d\n",a*b*c);
                return;
            }
        }
    }
}

int main()
{
    show();
    return 0;
}
View Code
Answer:31875000
Completed on Wed, 24 Jul 2013, 08:53
 

 10、Summation of primes

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

Find the sum of all the primes below two million.

10以下的质数的和是2 + 3 + 5 + 7 = 17.

找出两百万以下所有质数的和。

#include<stdio.h>
#include<math.h>
#include<stdbool.h>

#define N 2000000

bool prim(int n)
{
    int i;
    for(i=2; i*i<=n; i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

int main()
{
    int i;
    long long sum=2;
    for(i=3; i<=N; i=i+2)
    {
        if(prim(i))
        {
            sum+=i;
        }
    }
    printf("%lld\n",sum);

    return 0;
}
View Code
Answer:142913828922

Completed on Tue, 23 Jul 2013, 17:02

 

posted @ 2014-04-08 22:26  wuyudong  阅读(658)  评论(0)  编辑  收藏  举报
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