大数字相加

**

• 如 78789890000000098978加5647638738398； 》方法
  先都倒过来数对位相加，进位保留十位数，不进化为0；最后再将得数倒回来即可；（有可能两数位数不等就要加到位数少的，在单独存较多位的数；！！！进阶很需要思考！！！）**

  #include<stdio.h>
  #include<string.h>
  char a[1000];
  char b[1000];
  char c[1001];
  void fun(char a[], int n)
  {
  int i, t = 0;
  char c;
  n = n-1;
  for (i = n, t; i > t; i–, t++)
  {
  c = a[i];
  a[i] = a[t];
  a[t] = c;
  }
  }
  int main()
  {
  int s1, s2, s3, max, z, t, i = 0, x, y;
  gets_s(a);
  gets_s(b);
  s1 = strlen(a);
  s2 = strlen(b);
  fun(a, s1);倒过来数
  fun(b, s2);倒过来数
  if (s1 > s2)
  max = s1;
  else
  max = s2;
  z = 0;
  do
  {
  x = a[i] - ‘0’;
  y = b[i] - ‘0’;
  z = z + x + y;
  if (z >= 10)
  {
  t = z % 10;
  z = z / 10;
  c[i] = t + ‘0’;
  }
  else if (z<10)
  {
  c[i] = z + ‘0’;
  z = 0;
  }
  i++;
  } while (i<s1&&i<s2);
  //两数位数不等
  if (i != max)
  {
  if (s1 > s2)
  {
  do
  {
  x = a[i] - ‘0’;
  z = z + x;
  if (z >= 10)
  {z = z % 10;
  t = z / 10;
  c[i] = z + ‘0’;
  }
  else if (z < 10)
  {
  c[i] = z + ‘0’;
  z = 0;
  }
  i++;
  } while (i != max);
  }
  if (s2 > s1)
  {
  do
  {
  x = b[i] - ‘0’;
  z = z + x;
  if (z >= 10)
  {
  t = z / 10;
  z = z % 10;
  c[i] = z + ‘0’;
  }
  else if (z < 10)
  {
  c[i] = z + ‘0’;
  z = 0;
  }
  i++;
  } while (i != max);
  }
  }
  if (z != 0)
  {
  c[i] = z+‘0’;
  c[i + 1] = ‘\0’;
  }
  else
  c[i] = ‘\0’;
  s3 = strlen©;
  fun(c, s3);数倒回来
  puts©;
  return 0;
  }