高精度计算 大数乘法

#include <iostream>#include <cstring>using std::cin;using std::cout;using std::endl;char *multiplication( char *s1, char *s2 );int check( int *a, int n );//进位，去零，返回去零后的长度int *StrToInt(char *str);//将逆序字符串并转换成整型数组char *IntToStr( int *a, int n );//逆序整型数组并转换成字符串int main(){    char s1[100000],s2[100000];    //cin >> s1 >> s2;    while ( cin >> s1 >> s2 )        cout << multiplication(s1,s2) << endl;    return 0;}char *multiplication( char *s1, char *s2 ){    int len1 = strlen(s1), len2 = strlen(s2);    int len = len1 + len2;    int *t1 = new int [len1];    int *t2 = new int [len2];    int *t  = new int [len]();    t1 = StrToInt(s1);    t2 = StrToInt(s2);    int i = 0, j = 0;    for ( i = 0; i != len1; ++i )        for ( j = 0; j != len2; ++j )            t[i + j] += t1[i]*t2[j];    len = check(t, len);    return IntToStr(t, len);}//将逆序字符串并转换成整型数组int *StrToInt(char *str){    int len = strlen(str);    int *a = new int [len];    for ( int i = 0; i != len; ++i )        a[i] = str[len - i - 1] - '0';    return a;}//逆序整型数组并转换成字符串char *IntToStr( int *a, int n ){    char *str = new char [n + 1];    for ( int i = 0; i != n; ++i )        str[i] = a[n - i - 1] + '0';    str[n] = '\0';    return str;}//进位，去零，返回去零后的长度int check( int *a, int n ){    int i = 0;    for ( i = 0; i != n; ++i ){        if ( a[i] > 9 ){            a[i + 1] = a[i + 1] + a[i]/10;            a[i] %= 10;        }    }    while ( 0 == a[n - 1] && n > 1 )        --n;    return n;}/**优化过的函数,更加精简。int *Str2Int( char *s ){    int len = strlen(s);    int *a = new int [len];    int i(0);    while ( i != len )        a[i++] = s[len - 1 - i] - '0';    return a;}char *Int2Str( int *a, int len ){    char *s = new char [len + 1];    int i(0);    while ( i != len )        s[i++] = a[len - 1 - i] + '0';    s[len] = '\0';    return s;}int check( int *a, int len ){    for ( int i(0); i != len; ++i )        if ( a[i] > 9 ){            a[i + 1] += a[i]/10;            a[i] = a[i]%10;        }    //去零,确定长度    while ( 0 == a[len - 1] && len > 1)        --len;    return len;}*/