Question

C语言程序求1—100之间的奇数和和偶数和

Answer 1

int i, sum = 0;for(i = 1; i < 100; i +=2) sum += i;printf("1-100以内奇数和为%d", sum);sum = 0;for(i = 2; i <= 100; i += 2) sum += i;printf("1-100以内偶数和为%d", sum);

Answer 2

参考代码如下:
#include<stdio.h>

int main()
{
int i,s1=0,s2=0,s=0;
for(i=1;i<=100;++i){
if(i%2==0)
s2+=i;
else
s1+=i;
s+=i;
}
printf("奇数和%d,偶数和%d,总和%d",s1,s2,s);
return 0;
}

Answer 3

#include <stdio.h> void main(){ int i; int odd_sum; int even_sum; odd_sum = 0; even_sum = 0; for(i=1; i<100; i+=2) { odd_sum += i; even_sum += i+1; } printf("Odd Sum Is:%d,Even Sum Is:%d\n ",odd_sum,even_sum);}

Answer 4

Console.ReadLine();
是多的

Answer 5

两种方法功能参考
1、
#include "stdio.h"
int main()
{
int countou,countji=0;
int i=0;
for(i=1;i
2、
#include <stdio.h>
int main()
{
int i;
int ji=0, ou=0;
for (i=1;i<=100;i++) {
if (i%2 == 0)
ou += i;
else
ji += i;
}
printf("奇数之和：%d\n偶数之和：%d\n", ji, ou);
return 0;
}