小出一道题(乘法运算的程序)
描述从键盘读入个100以内的正整数,进行乘法计算并输出。
输入
多组测试数据，每组测试数据包括两个整数m,n仅一行，以0 0 结尾。两个数的第一位和最后一位都不是0.
输出
输出格式: 两个乘数的末位数对齐 乘号后面紧跟着第二个乘数,线的长度以最长的数的位数为准,每两组测试数据中间，输出一个空行。
样例输入
89 13
2 3
0 0
样例输出
89
x13
----
267
89
----
1157
2
x3
--
6
程序代码：#include <stdio.h>
#include <stdlib.h>
int main(void) {
int m, n, i = 0, j, k, inputs_size = 10, * inputs = (int *)malloc(sizeof(int) * inputs_size);
int m_width, n_width, m_n_width, m_n_10_width, result_width, temp;
while(scanf("%d%d", &n, &m) == 2 && n != 0 && m != 0) {
if(n > 100 || m > 100 || n < 0 || m < 0)
continue;
if(i >= inputs_size) {
inputs = (int *)realloc(inputs, inputs_size += 10);
}
inputs[i] = m;
inputs[i + 1] = n;
i += 2;
}
for(j = 0; j < i; j += 2) {
m = inputs[j];
n = inputs[j + 1];
m_width = 0, n_width = 0, m_n_width = 0, m_n_10_width = 0, result_width = 0;
temp = m;
while(temp) {
temp /= 10;
m_width++;
}
temp = n;
while(temp) {
temp /= 10;
n_width++;
}
temp = n * (m % 10);
while(temp) {
temp /= 10;
m_n_width++;
}
temp = n * (m/ 10 % 10);
while(temp) {
temp /= 10;
m_n_10_width++;
}
temp = n * m;
while(temp) {
temp /= 10;
result_width++;
}
result_width = result_width < 2 ? 2 : result_width;
for(k = 0; k < result_width - m_width; k++)
printf(" ");
printf("%d\n", n);
for(k = 0; k < result_width - n_width - 1; k++)
printf(" ");
printf("x%d\n", m);
for(k = 0; k < result_width; k++)
printf("-");
printf("\n");
for(k = 0; k < result_width - m_n_width; k++)
printf(" ");
printf("%d\n", n * (m % 10));
if(n / 10 != 0) {
if(result_width == 4 && m_n_10_width < 3)
printf(" ");
printf("%d\n", n * (m / 10 % 10));
for(k = 0; k < result_width; k++)
printf("-");
printf("\n%d\n\n", n * m);
} else {
printf("\n");
}
}
return 0;
}
[ 本帖最后由 lz1091914999 于 2011-6-4 22:16 编辑 ]
