C语言编程，求100以内的自然数中奇数之和。

Question

如题所述

Answer 1

#include<stdio.h>

int main()

{int i,s=0;        //s为累加器，初值为0

for(i=1;i<100;i+=2)    //穷举100以内所有的奇数

s+=i;        //进行累加

printf("100以内的自然数中奇数之和=%d\n",s);    //输出结果

return 0;

}

Answer 2

//编译通过#include <stdio.h>
void main()
{
int nSum = 1;
for (int i=2; i<=100; i++)
{
if (i%2 != 0)
{
nSum += i;
}
}
printf("0~100的奇数之和为：%d\n", nSum);
}

Answer 3

#include <stdio.h>

int main(void)
{
int i = 1;
int sum = 0;
while(i <= 99) {
sum += i;
i += 2;
}
printf("%d", sum);
return 0;
}本回答被网友采纳

Answer 4

#include<stdio.h>
int main(void) {
int i=0;
int sum = 0;
for (i=0; i<=100; i++) {
if((i % 2) == 1) {
sum += i;
}
else {
sum = sum + 0;
}
}
printf("100以内的自然数奇数和为%d\n",sum);
return 0;
}

Answer 5

#include <stdio.h>void main()
{
int i,sum;
sum=0;
for(i=1;i<100;i+=2)
{
sum+=i;
}
printf("%d\n",sum);
}