大整数相加和大整数相乘

#include<iostream>
#include<string>
using namespace std;
//字符串倒置
void reverse(char *str)
{
    int len=strlen(str);
    for(int i=0;i<len/2;i++)
    {
        char temp=str[i];
        str[i]=str[len-i-1];
        str[len-i-1]=temp;
    }
}
//大数求和
void bignumsum(char * ope1,char * ope2,char * result)
{
    reverse(ope1);
    reverse(ope2);
    int len1=strlen(ope1);
    int len2=strlen(ope2);
    int max=len1;
    if(max<len2)
    {
        max=len2;
    }
    memset(result,'0',max+1);
    result[max+1]='';
    int acc=0;
    int i=0;
    while(i<len1 && i<len2)
    {
        int temp=ope1[i]-'0'+ope2[i]-'0'+acc;
        acc=temp/10;
        result[i]=temp%10+'0';
        i++;
    }
    if(i<len1)
    {
        while(i<len1)
        {
            int temp=ope1[i]-'0'+acc;
            acc=temp/10;
            result[i]=temp%10+'0';
            i++;
        }
    }
    if(i<len2)
    {
        while(i<len2)
        {
            int temp=ope2[i]-'0'+acc;
            acc=temp/10;
            result[i]=temp%10+'0';
            i++;
        }
    }
    //考虑到进位，所以result的长度有可能比ope1和ope2中最大的数据长度还多出一位
    if(acc)
    {
        result[i]=acc+'0';
    }
    else
    {
        result[i]='';
    }
    reverse(result);
}
int main()
{
    char num1[4]="456";
    char num2[5]="5678";
    char num3[6];
    bignumsum(num1,num2,num3);
    cout<<num3<<endl;

    char num4[4]="456";
    char num5[5]="678";
    char num6[6];
    bignumsum(num4,num5,num6);
    cout<<num6<<endl;

    char num7[4]="111";
    char num8[5]="2";
    char num9[6];
    bignumsum(num7,num8,num9);
    cout<<num9<<endl;
    return 0;
}

#include<iostream>
#include<string>
using namespace std;

void reverse(char *str)
{
    int len=strlen(str);
    for(int i=0;i<len/2;i++)
    {
        char temp=str[i];
        str[i]=str[len-i-1];
        str[len-i-1]=temp;
    }
}
void bignummultiply(char *ope1,char *ope2,char *result)
{
    reverse(ope1);
    reverse(ope2);
    int len1=strlen(ope1);
    int len2=strlen(ope2);
    memset(result,'0',len1+len2);//因为len1和len2两个长度的整数相乘结果最大为len1+len2
    result[len1+len2]='';
    int acc=0;//加法进位
    int mcc=0;//乘法进位
    for(int i=0;i<len2;i++)
    {
        acc=0;
        mcc=0;
        for(int j=0;j<len1;j++)
        {
            int temp1=(ope1[j]-'0')*(ope2[i]-'0')+mcc;
            mcc=temp1/10;
            temp1=temp1%10;
            int temp2=result[i+j]-'0'+temp1+acc;
            acc=temp2/10;
            result[i+j]=temp2%10+'0';
        }
        result[i+len1]=acc+mcc+'0';
    }
    //这里有一个去除后面0的程序，如78900将变成789，这样倒过来之后就会使987，而不是00987
    int k=len1+len2-1;
    while('0'==result[k] && k>=0)
    {
        result[k--]='';
    }
    //最后经result倒置回来得到最终结果
    reverse(result);
}
int main()
{
    char n1[5]="12";
    char n2[5]="2";
    char n3[10];
    bignummultiply(n1,n2,n3);
    cout<<n3<<endl;

    char n4[5]="9";
    char n5[5]="9";
    char n6[10];
    bignummultiply(n4,n5,n6);
    cout<<n6<<endl;

    char n7[5]="888";
    char n8[5]="66";
    char n9[10];
    bignummultiply(n7,n8,n9);
    cout<<n9<<endl;
    return 0;
}