高精度乘法

Problem 1: 高精度乘法

Time Limit:1 Ms|  Memory Limit:128 MB
Difficulty:15

Description

运算两个高精度数相乘的积(1<=两个数的位数<=1000)

Input

输入 两个数

Output

乘积结果

Sample Input

987654321 123456789

Sample Output

121932631112635269

Hint

只有一组测试数据，进行多次测试

代码

#include<stdio.h>
#include<string.h>
#define MAXSIZE 1000

char m[MAXSIZE], n[MAXSIZE];
int m_1[MAXSIZE], n_1[MAXSIZE];
int s[MAXSIZE * MAXSIZE]; 

int main()
{
	int dig, k = 0 , t;
	int lenm, lenn, i, j;

	scanf("%s%s", n, m);
	lenm = strlen(m);
	lenn = strlen(n);
	for(i = 0; i < lenm; i++)
		m_1[i] = m[i] - 48;
	for(i = 0; i < lenn; i++)
		n_1[i] = n[i] - 48;
	for(j = lenm - 1; j >= 0; j--)				//先将各个数相乘所得结果保存在数组中
	{
		t = k;
		for(i = lenn - 1; i >= 0; i--)
		{
			s[t] += m_1[j] * n_1[i];
			t++;
		}
		++k;
		dig = t;
	}
	for(i = 0; i < dig; i++)
		while(s[i] >= 10)
		{
			s[i] -= 10;
			s[i + 1] += 1;
		}
	if(s[dig] != 0)
	{
		for(i = dig; i >= 0; i--)
			printf("%d", s[i]);
		printf("\n");
	}
	else
	{
		for(i = dig - 1; i >= 0; i--)
			printf("%d", s[i]);
		printf("\n");
	}
	return 0;
}