OJ中处理超大数据的方法

处理大数时如果用整数法表示，就算是long int范围是-2^63~2^63-1，如果输入上千位，还是会挂掉，浪费空间。

所以说要用字符串的形式进行处理！

比如这道OJ

/*Sum of Numbers

Write a program which reads an integer and prints sum of its digits.

Input
The input consists of multiple datasets. For each dataset, an integer x is given in a line. The number of digits in x does not exceed 1000.

The input ends with a line including single zero. Your program should not process for this terminal symbol.

Output
For each dataset, print the sum of digits in x.*/

一开始用整数法，挂掉

#include <stdio.h>
#include <iostream>
using namespace std;

int main(){

 while(true){
  long long int a;
  int r, s=0;
  cin>>a;
  if(a==0)
   break;
  else{
   while(a>0){
    r = a % 10;
    s = s + r;
    a = a / 10;
   }
   
   cout<<s<<endl;
  }
 }
 cout<<endl;
 
 return 0;
}

换用字符串的方法，A了

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

int main(){

 char a[1001];
 while(cin>>a, strcmp(a,"0")){ // 输入为0时结束程序 
  int n = 0;
  for(int i=0;i<strlen(a);i++) //将字符转换成整数 
   n += a[i] - '0';
  
  cout<<n<<endl;
 } 

 return 0;
}

注意一下字符串怎么处理输入结束标识符的