SDNU 1232.A*B Problem（高精度）

这道题虽然数据很大，但是可以用数学的乘法运算来解决

Description

Calculate a*b

Input

Input contains multiple test cases.

Each test case contains two number a and b (0<=a,b<=10^1000),

Output

Output a*b

Sample Input

12345678987654321 98765432123456789

Sample Output

1219326320073159566072245112635269

Hint

练习模拟大数乘法

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>

using namespace std;

string a, b;
int x[1000+8], y[1000+8], sum[1000000+8], t;
int main()
{
    while(cin >> a >> b)
    {
        int lena = a.size();
        int lenb = b.size();
        int lens = 0;
        memset(sum, 0, sizeof(sum));
        for(int i = 0; i<lena; i++)
            {
                x[lena-i] = a[i]-48;//-48等于-'0',并且把这个字符串倒序输入数列X中
            }
        for(int i = 0; i<lenb; i++)
        {
            y[lenb-i] = b[i]-48;
        }
        for(int i = 1; i<=lena; i++)//像数学乘法一样，把所有的数都乘起来
        {
            t = 0;//记录要进的位数
            for(int j = 1; j <= lenb; j++)
            {
                sum[i+j-1] += x[i]*y[j]+t;
                t = sum[i+j-1]/10;
                sum[i+j-1] %= 10;
            }
            sum[i+lenb] = t;
        }
        lens = lena+lenb;
        while(sum[lens] == 0 &&lens>1) lens--;
        for(int i = lens; i >= 1; i--)printf("%d", sum[i]);
        printf("
");
    }
    return 0;
}