方法还是与高精度加法的方法差不多,列一个竖式,就可以发现其中的规律了。
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 3001
using namespace std ;
int main ( )
{
char a1 [ N ] , b1 [ N ] ;
int a [ N ] , b [ N ] , c [ N ] , x ;
memset ( a , 0 , sizeof ( a ) ) ;
memset ( b , 0 , sizeof ( b ) ) ;
memset ( c , 0 , sizeof ( c ) ) ;
gets ( a1 ) ;
gets ( b1 ) ;
int lena = strlen ( a1 ) , lenb = strlen ( b1 ) ;
for ( int i = 0 ; i < lena ; i ++ ) a [ lena - i ] = a1 [ i ] - 48 ;
for ( int i = 0 ; i < lenb ; i ++ ) b [ lenb - i ] = b1 [ i ] - 48 ;
for ( int i = 0 ; i <= lena ; i ++ )
{
x = 0 ; //用来存放进位
for ( int j = 1 ; j <= lenb ; j ++ )
{
c [ i + j - 1 ] += a [ i ] * b [ j ] + x ;
x = c [ i + j - 1 ] / 10 ;
c [ i + j - 1 ] %= 10 ;
}
c [ i + lenb ] = x ; //进位
}
int lenc = lena + lenb ;
while ( c [ lenc ] == 0 && lenc > 1 ) lenc -- ; //删除前导0
for ( int i = lenc ; i > 0 ; i -- ) printf ( "%d" , c [ i ] ) ;
return 0 ;
}
相关链接:
C++模板小站:
https://blog.csdn.net/ZJ_MRZ/article/details/80950647
C++快速幂模板:
https://blog.csdn.net/zj_mrz/article/details/80950616
C++高精度加法模板:
https://blog.csdn.net/zj_mrz/article/details/80948327
C++高精度减法模板:
https://blog.csdn.net/zj_mrz/article/details/80965324