http://projecteuler.net/problem=4 【参考解法】

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本文探讨了如何通过翻转特定数值并寻找满足特定条件的三位数，涉及数值处理和数学逻辑。

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#include <iostream>

using std::cout;
using std::endl;

typedef unsigned int uint32;

const uint32 MIN = 99;
const uint32 MAX = 999;

uint32 revers_num(uint32 k)
{
    uint32 rk = 0;
    while( k )
    {
        rk = 10 * rk + k % 10;
        k /= 10;
    }

    //cout << "k=" << k << ", rk=" << rk << endl;
    return rk;
}

bool is_3digit_num( uint32 n)
{
    bool ret = false;
    for(uint32 i = MAX; i != MIN; --i )
    {
        int t = n / i;
        if ( t * i == n && t > MIN && t <= MAX )
        {
            cout << "t=" << t << ", i=" << i << endl;
            ret = true;
            break;
        }
    }

    return ret;
}

uint32 resolveP4()
{
    uint32 result = 0;
    for( uint32 j = MAX; j != MIN; --j )
    {
        uint32 reverse = revers_num( j );
        uint32 n = j * (MAX + 1) + reverse;
        if ( is_3digit_num ( n ) )
        {
            result = n;
            break;
        }
    }

    return result;
}

int main()
{
    //uint32 result = revers_num(800);
    uint32 result = resolveP4();
    cout << "result = " << result << endl;

    return 0;
}
/*当我解决之后，发现了一个论坛上有人分享了一个更好的解决方法：
The palindrome "abccba" can be written as
11(9091a + 910b + 100c) = mn;
a,b & c being 1 digit integers and m & n being 3 digit intergers.

Let 11 * 10 < m < 11 * 90;

for(int a=9; a>=1; a--)
  for(int b=9; b>=0; b--)
    for(int c=9; c>=0; c--){
      num = 9091 * a + 910 * b + 100 * c;
      for(int divider=90; divider>=10; divider--){
        //look for divider that can divide 
        //and also doesn't make n > 999
	if((num % divider) == 0){
	  if((num / divider) > 999)
	    break;
	  else
	    result = num * 11; //Found it!
	} else { break; }
      } */