c++编程，求100之内的自然数中奇数之和

Question

（用do,dowhile,for分别编写）
是c++6.0不是7.0对不起刚忘说了

Answer 1

//用do while语句
#include<stdio.h>
int main()
{
int i=1;
int sum=0;
do{
sum+=i;
i+=2;
}while(i<=99);
printf("100以内奇数和为sum=%d\n",sum);
return 0;
}

//用for语句
#include<stdio.h>
int main()
{
int i,n;
int sum=0;
for(i=1;i<=50;i++){
n=2*i-1;
sum=sum+n;
}
printf("100以内奇数和为sum=%d\n",sum);
return 0;
}

//用while语句
#include"stdio.h"
int main()
{
int i=1,n;
int sum=0;
while(i<=100)
{
if(i%2==0)//判断奇偶性
sum=sum+(i-1);
i++;
}
printf("100以内奇数和为sum=%d\n",sum);
return 0;
}本回答被网友采纳

Answer 2

int i=1;
int sum=0;
do
{
sum+=i;
i+=2;

}while(i<=99)

Answer 3

# include <stdio.h>
main()
{int n=0,i;
for(i=1;i<=100;i++)
if(i%2!=0)
n=n+i;
printf("the resule is %d",n);

}

#include<stdio.h>
main()
{int n=0,i=1;
do
{if(i%2!=0)
n=n+i;
i++;
}while(i<=100);
printf("the result is %d",n);

}

#include<stdio.h>
main()
{int n=0,i=1;
while(i<=100)
{if(i%2!=0)
n=n+i;
i++;
}
printf("the result is %d",n);

}本回答被网友采纳

Answer 4

//判断是否为奇数
int sum=0;
for(int i=1;i<100;i++)
{
if(i%2==0)
sum+=i;
}
return sum

int sum=0,i=0;
do
{
if(i%2==0)
sum=sum+i;
i++;
}while(i<=100)
return sum;

Answer 5

....还用3个分开写