projecteulerProblem(1-10)

地址在下面，都是些有趣的数学题，有兴趣的可以去玩玩.
http://www.projecteuler.net
呵呵我刚做完前10题，以后每做完10题都会发到我博客上。
PS:我用的是python和c写的。
1

引用

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

def find():
    return reduce(lambda x,y:x+y,[x for x in xrange(3,1000)  if  not x%3 or not x%5])
print find()

2

引用

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

Find the sum of all the even-valued terms in the sequence which do not exceed one million.

import itertools
def Fibonacci():
    x, y = 1,2
    while True:
        if x%2==0:
            yield x
        x, y = y, x + y
print reduce(lambda x,y:x+y,itertools.takewhile(lambda x:x<1000000,Fibonacci()))

3

引用

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 317584931803?

#include <stdio.h>
void main()   
{   
  __int64 n=317584931803;
  __int64 i;
    for(i=2;i<=n;i++)   
    {   
    while(n!=i){   
      if(n%i==0){ 
        n=n/i;   
      }   
      else   
        break; 
    }   
    }   
    printf("%d",n);   
}

4

引用

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91  99.

Find the largest palindrome made from the product of two 3-digit numbers.

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

void main( )
{
  char xy[7];
  int str,half,flag;
  for(int i=998001;i>10000;i--){
    itoa(i,xy,10);
    str=strlen(xy);
    half=str/2;
    for(int j=0;j<half;j++){
      if(xy[j]!=xy[--str])
        break;
      else if(j==half-1)
        for(int k=100;k<1000;k++){
          int tmp=i/k;
          int tmp2=i%k;
          if(tmp2==0&&99<tmp&&tmp<1000){
            printf("%d=%d*%d",i,tmp,k);
            flag=1;
            break;
          }
        }
    }
    if(flag==1)
      break;
  }

}

5

引用

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?

#include <stdio.h>
int pub(int i,int j);
void main(){
	int result=2;
	int i=3;
	while(true){
		if(i>19)
			break;
		int tt=pub(result,i);	
		result=result*i/tt;
		i+=1;
	}
	printf("%d\n",result);
}

int pub(int i,int j)
{
	int temp;
	if(i<j){
		int tmp=i;
		i=j;
		j=tmp;
	}
	while(true){
			temp=i%j;
			if(temp==0)
				break;
			else{
				i=j;
				j=temp;
			}
	}
	return j;
}

6

引用

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.

def deff():
    return  sum(xrange(1,101))**2-reduce(lambda x,y: x+y,[x**2 for x in xrange(1,101)])
print deff()

7

引用

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?

import itertools
def pre( ):
    D = {  }
    yield 2
    for q in itertools.islice(itertools.count(3), 0, None, 2):
        p = D.pop(q, None)
        if p is None:
            D[q*q] = q
            yield q
        else:
            x = p + q
            while x in D or not (x&1):
                x += p
            D[x] = p
print list(itertools.islice(pre(),100001))[10000]

8

引用

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

def find():
    a="""73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
"""
    result=1
    f=a.replace("\n","")
    for i in xrange(0,len(f)-4):
        tmp= reduce(lambda x, y: int(x)*int(y), f[i:i+5])
        if result<tmp:
            result=tmp
    return result
print find()

9

引用

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.

#include <stdio.h>
void main()
{
	for(int i=2;i<1000;i++)
	{
		for(int j=i+1;j<1000;j++)
			if(i*i+j*j==(1000-i-j)*(1000-i-j))			
				printf("%d\n",i*j*(1000-j-i));	
	}
}

10

引用

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

Find the sum of all the primes below one million.

import itertools
def pre( ):
    D = {  }
    yield 2
    for q in itertools.islice(itertools.count(3), 0, None, 2):
        p = D.pop(q, None)
        if p is None:
            D[q*q] = q
            yield q
        else:
            x = p + q
            while x in D or not (x&1):
                x += p
            D[x] = p
print sum(list(itertools.takewhile(lambda x:x<1000000,pre())))

评论

1 楼 Eastsun 2007-05-21  

唔,貌似楼主没有考虑运行效率的问题~