大数加法~

描述

给定任意位数长度的两个数，求它们数值之和。

输入

每个测试用例含有两个数，代表两个任意位数长度的数A，B.

输出

计算出这两个任意位数长度的数之和。

样例输入

123451234512345   543215432154321
12.345    54.321

样例输出

666666666666666
66.666

#include<iostream>
#include<cstring>
using namespace std;
int la, lb;
char a[1000000],b[1000000];
//数组补0
void add0(char q[],int t,int j)
{
    int i;
    for(i = 0; i < j; i++)
    {
        q[t++] = ‘0‘;
    }
    q[t] = ‘\0‘;
}
int main()
{
    int as, bs, diana, dianb, i, j;
    char *pa, *pb, ch;
    while(~scanf("%s %s",a,b))
    {
        pa = strchr(a,‘.‘);
        pb = strchr(b,‘.‘);
        la = strlen(a);
        lb = strlen(b);
        if(pa == NULL)  //如果输入的a是整数那么最后加小数点
        {
            a[la] = ‘.‘;
            la = la+1;
            pa = strchr(a,‘.‘);
        }
        if(pb == NULL)    //如果输入的a是整数那么最后加小数点
        {
            b[lb] = ‘.‘;
            lb = lb+1;
            pb = strchr(b,‘.‘);
        }
//    if(pa!=NULL)
//    {
           diana = pa-a;
           dianb = pb-b;
            j = la-diana-lb+dianb;
            add0(b,lb,j);
//    }
//     if(pb!= NULL)
//     {
//          diana = pa-a;
//          dianb = pb-b;
            j = lb-dianb-la+diana;
            add0(a,la,j);
 //       }
        strrev(a);
        strrev(b);
        la = strlen(a);
        lb = strlen(b);
        if(la>lb)
            add0(b,lb,la-lb);
        else if(la<lb)
            add0(a,la,lb-la);//printf("%s %s\n",a,b);
        j = 0;
        la = strlen(a);
        for(i = 0; i < la; i++)
        {
            if(a[i]==‘.‘)
                continue;
            ch = (a[i]+b[i]-96+j)%10+‘0‘;
            j = (a[i]+b[i]-96+j)/10;
            a[i] = ch;
        }
        a[i] = ‘\0‘;
        strrev(a);
        la = strlen(a);
        i = la-1;
        while(a[i]==‘0‘)
        {
            i--;
        }
        a[i+1] = ‘\0‘;
        if(a[i]==‘.‘)
            a[i] = ‘\0‘;
        if(j)
            printf("%d",j);
        printf("%s\n",a);
    }
    return 0;
}