A + B Problem (Big integer version) 【高精度】

高精度

Time Limit: 1000 MS Memory Limit: 65536 KB

链接：https://www.oj.swust.edu.cn/problem/show/1202

Description

Input and output are the same with problem 1001.
 But A and B are big non-negative integers.
The biggest integer is less than 10^500.

Input

Input contains multiple test cases. 
Each case have two big non-negative integers a,b.

Output

For each case,Output a+b.

Sample Input

1234567890987654321
9876543210123456789
123456789123456789
321654987321654987
9081321110693270343633073697474256143563558458718976746753830538032062222085722974121768604305613921745580037409259811952655310075487163797179490457039169594160088430571674960498834085812920457916453747019461644031395307920624947349951053530086146486307198155590763466429392673709525428510973272600608981219760099374675982933766845473509473676470788342281338779191792495900393751209539300628363443012
6538005862664913074813656220643842443844131905754565672075358391135537108795991638155474452610874309742867231360502542308382199053675592825240788613991898567277116881793749340807728335795394301261629479870548736450984003401594705923178314906195914825136973281314862289454100745237769034410057080703111299605127114594552921209928891515242515620324828055912854227507525717981351447473570262981491527798

Sample Output

11111111101111111110
445111776445111776
15619326973358183418446729918118098587407690364473542418829188929167599330881714612277243056916488231488447268769762354261037509129162756622420279071031068161437205312365424301306562421608314759178083226890010380482379311322219653273129368436282061311444171436905625755883493418947294462921030353303720280824887213969228904143695736988751989296795616398194193006699318213881745198683109563609854970810

Hint

Note that there are leading zeros!

完整代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[505],b[505],sum[505];
char s1[505],s2[505];
int main()
{
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(sum,0,sizeof(sum));
        int l1=strlen(s1);
        int l2=strlen(s2);
        for(int i=0;i<l1;i++)
            a[i]=s1[l1-1-i]-'0';//将s1里的数倒序存入a数组中
        for(int i=0;i<l2;i++)
            b[i]=s2[l2-1-i]-'0';//将s2里的数倒序存入a数组中
        int l=(l1>l2)?l1:l2;
        for(int i=0;i<l;i++)
        {
            sum[i]+=(a[i]+b[i]);//求和
            if(sum[i]>=10)//如果有进位
            {
                sum[i]-=10;//该位数减10
                sum[i+1]+=1;//下一位加1
            }
        }
        if(sum[l]>0)//如果最后的有进位，则再加一个长度
            l++;
        for(int i=l;i>0;i--)//去前导零
        {
            if(sum[i]==0)
                l--;
            else
                break;
        }
        for(int i=l;i>=0;i--)//输出该数
            printf("%d",sum[i]);
        printf("\n");
    }
    return 0;
}