大数乘法 - 大数作为字符串的C语言操作

【问题描述】

编写程序实现两个超长整数（大于等于0，每个最长80位数字）的乘法运算。

【输入形式】

从键盘分行读入两个超长整数，要考虑输入高位可能为0的情况（如00083），每行的最后都有回车换行。

【输出形式】

输出只有一行，是两个长整数的乘法运算结果，从高到低依次输出各位数字，各位数字紧密输出。除非结果为0，否则最高位不能为0。

【输入样例】

134098703578230056

002340980000000000

【输出样例】

313922383102564996494880000000000

【样例说明】

进行两个整数乘法运算，134098703578230056 * 002340980000000000 = 313922383102564996494880000000000。

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

void reverse(char str[])
{
int l=0,r=strlen(str)-1;
char tmp;
while (l < r)
{
tmp = str[l];
str[l] = str[r];
str[r] = tmp;
l++, r--;
}
}

void add(char A[],char B[], char rst[])
{
reverse(A);
reverse(B);
int i = 0;
int tmp, carryover = 0;
while (A[i] != '\0'&&B[i] != '\0')
{
tmp = A[i] - '0' + B[i] - '0';
if (tmp > 9)
{
tmp = tmp - 10;
rst[i] = tmp + '0';
if (carryover) rst[i]++;
carryover = 1;
}
else
{
rst[i] = tmp + '0';
if (carryover) rst[i]++;
carryover = 0;
}
i++;
}

if (B[i] == '\0' && A[i] != '\0')
{
rst[i] = A[i];
if (carryover) rst[i]++;
i++;
while (A[i] != '\0')
{
rst[i] = A[i];
i++;
}
}
else if (A[i] == '\0'&&B[i] != '\0')
{
rst[i] = B[i];
if (carryover) rst[i]++;
i++;
while (B[i] != '\0')
{
rst[i] = B[i];
i++;
}
}
else if (A[i] == '\0'&&B[i] == '\0'&&carryover)
{
rst[i] = 1;
i++;
}

while (rst[i-1] == '0'&&i>1) i--;
rst[i] = '\0';
reverse(A);
reverse(B);
if(A!=rst) reverse(rst);
}

void multiply(char A[], char B[], char rst[])
{
int i,j,k,carryover;
char intermed[161];
rst[0] = '0';rst[1]='\0';
int tmp;

reverse(A);
reverse(B);
i = 0;
carryover = 0;
while (B[i] != '\0')
{
for (j = 0; j < strlen(A); j++)
{
tmp = (A[j] - '0')*(B[i] - '0');
intermed[j] = '0' + tmp % 10;
intermed[j] += carryover;
carryover = 0;
if (intermed[j] > '9')
{
intermed[j] -= 10;
carryover++;
}
while (tmp > 9)
{
tmp -= 10;
carryover++;
}
}
if (carryover) intermed[j++] = '0' + carryover;
carryover = 0;
intermed[j] = '\0';
reverse(intermed);
for (k = 0; k < i; k++)	intermed[j+k] = '0';
intermed[j+k] = '\0';

add(rst, intermed, rst);
i++;
}
reverse(A);
reverse(B);
}

int main()
{
char A[81],B[81];
char rst[161];
scanf("%s",A);
scanf("%s",B);
multiply(A, B, rst);
printf("%s",rst);
return 0;
}