c语言作业

(1)奇数累加 s = 1 + 3 + 5 + 7 +...+ 99

(2)偶数累加 s = 2 + 4 + 6 + 8 +...+ 100

（1）s = 1 + 3 + 5 + 7 + ... + 99;

#include <stdio.h> 
 int main(){ 
  int s = 0 ;
  int i = 1 ;
  
  while(i <= 100){
          printf("%d \n",i); 
          s = s + i; i+=2; 
}   
  printf("1 + 3 + 5 + 7 + ... + 99的和为：%d \n",s);
  return 0; 
}

（2）s = 2 + 4 + 6 + 8 + ... + 100;

#include <stdio.h> 
 int main(){ 
  int s = 0 ; 
  int i = 2 ;
  
 while(i <= 100){ printf("%d \n",i);
         s += i; 
         i +=2;
 } 
         printf("2 + 4 + 6 + 8 + ... + 100的和为：%d \n" , s);
  return 0;
}

运行如下：