c - 欧拉计划＃8

我试着自己解决这个问题，但过了一会儿就卡住了，所以看了看解决方案。这就是我得到的。

#include <stdio.h>

int main(void)
{
  char str[] =
    "73167176531330624919225119674426574742355349194934"
    "96983520312774506326239578318016984801869478851843"
    "85861560789112949495459501737958331952853208805511"
    "12540698747158523863050715693290963295227443043557"
    "66896648950445244523161731856403098711121722383113"
    "62229893423380308135336276614282806444486645238749"
    "30358907296290491560440772390713810515859307960866"
    "70172427121883998797908792274921901699720888093776"
    "65727333001053367881220235421809751254540594752243"
    "52584907711670556013604839586446706324415722155397"
    "53697817977846174064955149290862569321978468622482"
    "83972241375657056057490261407972968652414535100474"
    "82166370484403199890008895243450658541227588666881"
    "16427171479924442928230863465674813919123162824586"
    "17866458359124566529476545682848912883142607690042"
    "24219022671055626321111109370544217506941658960408"
    "07198403850962455444362981230987879927244284909188"
    "84580156166097919133875499200524063689912560717606"
    "05886116467109405077541002256983155200055935729725"
    "71636269561882670428252483600823257530420752963450";
  size_t len = sizeof str - 1;
  size_t i;
  unsigned max = 0;

  for (i = 0; i < len-4; i++) {
    unsigned p = 1;
    size_t j;

    for (j = 0; j < 5; j++) {
      p *= (unsigned)(str[i+j]-'0');
    }
    if (p > max) {
      max = p;
    }
  }
  printf("%u\n", max);
  return 0;
}

我的问题：我不明白为什么这一行有-'0'

 p *= (unsigned)(str[i+j]-'0');

对不起，如果我的问题很愚蠢。但我好像想不出来。

最佳答案

'0' - '0' == 0
'1' - '0' == 1
'2' - '0' == 2
etc.

基本上，它是用来把achar转换成相应的数字。
一种解释是字符直接映射到ASCII值：

'0' == 48
'1' == 49
'2' == 50
etc.

这些值是连续的。因此，当您减去最低的数字时，您将得到该数字之后的位置，这也是该数字的整数值。

关于c - 欧拉计划＃8，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/20710051/