6、

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：

1

² + 2

² + ... + 10

² = 385

前十个自然数的和的平方是：

(1 + 2 + ... + 10)

² = 55

² = 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 
      
       #include 
       
        #include 
        
         #include 
         
            #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;}
         
        
       
      

Answer:25164150

Completed on Tue, 2 Apr 2013, 06:57

 

---

 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 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 
      
       #include 
       
        #include 
        
         #include 
         
            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;}
         
        
       
      

Answer:104743

Completed on Thu, 4 Apr 2013, 17:34

 

---

 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

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 
      
       #include 
       
        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
       
      

Answer:40824

Completed on Sun, 17 Nov 2013, 11:16

 

---

 9、

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

a

² + 

b

² = 

c

²

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：

a

² + 

b

² = 

c

²

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
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            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;}
           
          
         
        
       
      

Answer:31875000

Completed on Wed, 24 Jul 2013, 08:53

 

---

 10、

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
      
       #include
       
        #include
        
         #define N 2000000bool 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;}
        
       
      

Answer:142913828922

Completed on Tue, 23 Jul 2013, 17:02