The bored Three-God get another boring question.
This problem is ask you plus two big nubmer, please help him, when you solve this problem you can
speak to Three-God,"Oh, Please give me a diffculte one, this one is too easy".
Input
There are muti-case
For each case, there are two integers, n, m (0 < n, m < 10^10000).
Output
Calculate two integers' sum
Sample Input
1 1
2 3
10000 100000
Sample Output
2
5
110000
题意 给两数 输出和 就是大数加法
因为数据小就直接暴力了,输入貌似会有前导0,输出也要前导0;本来写了超时了 再看才发现scanf前没加‘~’,要仔细= =
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
char a[10005];
char b[10005];
int c[10005];
int d[10005];
int e[10005];
int main()
{
while(~scanf("%s %s",a,b))
{
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
memset(e,0,sizeof(e));
int la=strlen(a);
int lb=strlen(b);
int lc=max(la,lb);
for(int i=0;i<la;i++)
c[i]=a[la-1-i]-'0';
for(int i=0;i<lb;i++)
d[i]=b[lb-1-i]-'0';
int k=0;
for(int i=0;i<lc;i++)
{
e[i]=c[i]+d[i]+k;
k=e[i]/10;
e[i]=e[i]%10;
}
if(k)
{
e[lc]=k;
lc++;
}
for(int i=lc-1;i>=0;i--)
{
printf("%d",e[i]);
}
printf("\n");
}
return 0;
}