C 解法, 执行用时: 1ms, 内存消耗: 376KB, 提交时间: 2020-07-18

#include <stdio.h>
#include <string.h>
void GetMulti(char s1[], char s2[])
{
    int num1[1000], num2[1000];
    int res[1000] = {0};
    for (int i = 0; i < strlen(s1); ++i) {
        num1[strlen(s1) - i - 1] = s1[i] - '0';
    }
    for (int i = 0; i < strlen(s2); ++i) {
        num2[strlen(s2) - 1 - i] = s2[i] - '0';
    }
    for (int i = 0; i < strlen(s1); ++i) {
        for (int j = 0; j < strlen(s2); ++j) {
            res[i + j] += num1[i] * num2[j];
        }
    }
    int tmp = 0;
    for (int i = 0; i <= strlen(s1) + strlen(s2) - 1; ++i) {
        res[i] += tmp;
        tmp = res[i] / 10;
        res[i] = res[i] % 10;
    }
    tmp = strlen(s1) + strlen(s2) - 1;
    while (res[tmp] == 0) {
        --tmp;
    }
    for (int i = tmp; i >= 0; --i) {
        printf("%d", res[i]);
    }
    printf("\n");
    return;
}
int main()
{
    char str1[1000], str2[1000];
    gets(str1);
    gets(str2);
    GetMulti(str1, str2);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 228KB, 提交时间: 2019-04-12

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

int main(int argc, char *argv[])
{
  int j, k, z;

#define SIZE 1000
  char str[SIZE+1];
  scanf("%s", str);

  int int1[SIZE];
  int number1;
  number1 = strlen(str);
  for (j = 0; j < number1; j++)
  {
    if (str[j] != '0') break;
  }
  for (k = j, z = 0; k < number1; k++, z++)
  {
    int1[z] = str[k] - '0';
  }
  number1 = z;

  int int2[SIZE];
  int number2;
  scanf("%s", str);
  number2 = strlen(str);
  for (j = 0; j < number2; j++)
  {
    if (str[j] != '0') break;
  }
  for (k = j, z = 0; k < number2; k++, z++)
  {
    int2[z] = str[k] - '0';
  }
  number2 = z;

  int resPos;
  int int3[SIZE+SIZE];
  int endResPos = sizeof(int3)/sizeof(int);
  for (k = 0; k < endResPos; k++) { int3[k] = 0; }

  for (k = 1; k <= number2; k++)
  {
    int numPos = k;
    int curNum = int2[number2 - k];
    for (j = 1; j <= number1; j++)
    {
      int x = curNum * int1[number1 - j];
      resPos = (endResPos-1) - (numPos-1) - (j-1);

      int3[resPos] += x;
      while (int3[resPos] >= 10)
      {
         int incr = int3[resPos]/10;
         int3[resPos] %= 10;
         resPos--;
         int3[resPos] += incr;
      }
    }
  }

  for (k = resPos; k < endResPos; k++)
  {
    printf("%d", int3[k]);
  }
  printf("\n");

  return 0;
}