使用C语言，求1到100之间的奇数和偶数之和，并输出，请高手指点

Question

如题所述

Answer 1

解法1:
#include <stdio.h>

int F1()
{
return 50*(1+99)/2;
}

int F2()
{
return 50*(2+100)/2;
}

void main()
{

printf("Odd number sum=%d,Even number sum=%d\n",F1(),F2());
}

解法2

#include <stdio.h>

int F1()
{
int i,sum=0;
for(i=1;i<=100;i+=2)
sum+=i;
return sum;
}

int F2()
{
int i,sum=0;
for(i=2;i<=100;i+=2)
sum+=i;
return sum;
}

void main()
{

printf("Odd number sum=%d,Even number sum=%d\n",F1(),F2());
}

Answer 2

搞定了啊

int main(void)
{
int s1=0;/*偶数和*/
s1=s2=0; /**奇数和/
for(i=1; i<=100; ++i)
{
if(i%2==0) s1 += i;
else s2 += i;
}
printf(" the even sum is %d\n", s1);
printf("the odd sum is %d\n", s2);
return 0;
}

Answer 3

#include <stdio.h>
int F1()
{
int sum = 0;
for(int i=1;i<=100;i++)
if(i % 2 == 0) //偶数
sum+=i
return sum;
}

int F2()
{
int sum=0;
for(int i=1;i<=100;i++)
if(i%2 == 1) //奇数
sum += i;
return sum;
}

void main()
{
printf("偶数和=%d,奇数和=%d\n",F1(),F2());
}

Answer 4

int i;
int jishu; 奇数
int oushu; 偶数
for (i=1;i<=100;i++){
if(i%2==0){
oushu=oushu+i;}
else{
jishu=jishu+i;}
}
最后输出就行了