编写程序，求1~100之间的奇数和偶数之和，并加以输出。用C语言

Answer 1

#include "stdio.h"
int main()
{
int countou,countji=0;
int i=0;
for(i=1;i<=100;i++)
{
if(i%2==0) countou+=i;
else countji+=i;
}
printf("偶数和:%d\n 奇数和:%d",countou,countji);
return 0;
}

Answer 2

既然是这样,那用得着那么复杂吗,直接从1加到100不就得了,
#include<stdio.h>
void
main()
{
int
a=1,b=1;
while(a<100)
{
a++;
b=b+a;
}
printf("%d\n",b);
}

Answer 3

#include<stdio.h>
main()
{
int sum1=0,sum2=0,k=2,i=1;
while(i<=99)
{
sum1=sum1+i;
i=i+2;
}
while (k<=100)
{
sum2=sum2+k;
k=k+2;
}
printf("奇数项和为%d ,偶数项和为%d",sum1,sum2)
;
}

Answer 4

既然是这样,那用得着那么复杂吗,直接从1加到100不就得了,
#include<stdio.h>
void main()
{
int a=1,b=1;
while(a<100)
{
a++;
b=b+a;
}
printf("%d\n",b);
}

Answer 5

countou也要初始化为0的

#include <stdio.h>
int main()
{
int countou=0,countji=0;
int i=0;

for(i=1;i<=100;i++)
{
if(i%2==0) countou+=i;
else countji+=i;
}
printf("偶数和:%d\n 奇数和:%d\n",countou,countji);
return 0;
}