sums



Description

Sometimes Ziwen need to operate even larger numbers. A limit of 1000 digits is so small… You have to find the sum of two numbers with maximal size of 1 000 000 digits.

Input

The first line contains a single integer N that is the length of the given integers(1 ≤ N ≤ 1 000 000). It is followed by these integers written in columns. That is, the next N lines contain two digits each, divided by a space. Each of the two given integers is not less than 0, and the length of their sum does not exceed N. The integers may contain leading zeroes.

Output

Output exactly N digits in a single line representing the sum of these two integers.

Sample Input

40 44 26 83 7

Sample Output

4750

代码

#include<iostream>using namespace std;main(){ int n; while(scanf("%d",&n)!=EOF){  int a[n][2];  for(int i=0;i<n;i++){   scanf("%d%d",&a[i][0],&a[i][1]);  }  int c[n+1];  for(int i=0;i<n+1;i++){   c[i]=0;  }  static int k=0;    for(int i=n-1;i>=0;i--){        c[i+1]=(a[i][0]+a[i][1]+k)%10;     k=(a[i][0]+a[i][1]+k)/10;    }    if(c[0]!=0) printf("%d",c[0]);    for(int i=1;i<n+1;i++){     printf("%d",c[i]);    }   }} 

算法还是较简单的，不过在开始测试，发现输出的总是00000X，后面才发现少了using namespace std； 但程序可以运行，不过无法的出正确的结果，原因待究。