这三道题都属于比较基础简单的题目，按照每道题的要求写就可以了。
1000：
Description
Your task is to Calculate a + b.
Input
Input contains multiple test cases. Each test case consists of a pair of integers a and b ( 0 <= a, b <= 20 ), separated by a space, one pair of integers per line.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample Input
1 1
Sample Output
2
代码如下：
#include<stdio.h>
int main()
{
int a, b;
while((scanf("%d%d", &a, &b)) != EOF)
printf("%d\n", a + b);
return 0;
}1001：
Description
Your task is to Calculate a + b.
Input
There are multiple test cases. Each test case contains only one line. Each line consists of a pair of real number a and b（0<=a,b<=1000000）, separated by a space.
Output
For each case, output the answer in one line rounded to 4 digits after the decimal point.
Sample Input
1 5 10 20 0.1 1.5
Sample Output
6.0000 30.0000 1.6000
代码如下：
#include<stdio.h>
int main()
{
double a, b;
while((scanf("%lf%lf", &a, &b)) != EOF)
{
printf("%.4lf\n", a + b);
}
return 0;
}1002：
Description
Your task is to Calculate a + b.
Input
There are multiple test cases. Each test case contains only one line. Each line consists of a pair of integers a and b ( 1 <= a, b<=1016 ) , separated by a space. Input is followed by a single line with a = 0, b = 0, which should not be processed.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample Input
1 5 10 20 10000000000000000 10000000000000000 0 0
Sample Output
6 30 20000000000000000
代码如下：
#include <iostream>
#include <math.h>
#include <stdlib.h>
using namespace std ;
long long int getNum(string s){
long long int num = 0 ;
int flag = 1 ;
if (s[0]=='-')
flag = -1 ;
for (int i = 0 ; i < s.length() ; ++i){
if (abs(s[i]-'4')<=5){
num+=(s[i]-'0') ;
num *= 10 ;
}
}
return flag*(num/10) ;
}
int main(){
string a ,b ;
while (cin>>a>>b){
if(getNum(a) != 0 && getNum(b) != 0){
cout<< getNum(a)+getNum(b)<<endl;
}
}
return 0;
}如果大家有更加简洁的代码，欢迎评论交流