补:==不能用来比较两个字符串是否相等,应该使用一个库函数—strcmp
#define _CRT_SECURE_NO_WARNINGS
#include
#include
方法1:
int main() {
int i = 0;
int j = 0;
int count = 0;
for (i = 101; i <= 200; i += 2)
{
for (j = 2; j < i; j++)
{
if (i%j == 0)
break;
}
if (i == j) {
count++;
printf("%d ", i);
}
}
printf("ncount=%d", count);
return 0;
}
优化:
#include
int main() {
int i = 0;
int j = 0;
int count = 0;
for (i = 101; i <= 200; i += 2) {
sqrt(i)可替换为i/2,得到的结果是一样的。
for (j = 2; j <= sqrt(i); j++)//i=a*b,其中,a和b至少有一个值<=根号i,这样就能只判断一半的j,实现代码的简便性。
{
if (i%j == 0)
break;
}
if (j > sqrt(i))//sqrt是数学开方的库函数—
{
count++;
printf("%d ", i);
}
}
printf("ncount=%d", count);
return 0;
}
方法1、
int main() {
int year = 0;
int count = 0;
for (year = 1000; year <= 2000; year++) {
if (year % 4 == 0 && year % 100 != 0)
{
printf(" %d", year);
count++;
}
else if (year % 400 == 0) {
count++;
printf(" %d", year);
}
}
printf("n其中有%d个闰年", count);
return 0;
}
方法2、
int main() {
int year = 0;
int count = 0;
for (year = 1000; year <= 2000; year++) {
if ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)) //“||”表示或者的意思,“&&”表示并且的意思。if分支条件可以进行嵌套
{
printf(" %d", year);
count++;
}
}
printf("n其中有%d个闰年", count);
return 0;
}
给定两个数,求这两个数的最大公约数
辗转相除法
(1)
int main() {
int x=0;
int y=0;
int i = 0;
scanf("%d%d", &x, &y);
for (i;; ){
if (x%y ==0) {
printf("公约数为%d", y);
break;
}
else
{
int z = x % y;
x = y;
y = z;
}
}
return 0;
}
写一个代码打印1~100之间所有3的倍数的数字
int main() {
int x = 0;
for (x = 1; x <= 100; x++)
{
if (x % 3 == 0)
printf("%dn", x);
}
return 0;
}
写三个数按从大到小输出
int main() {
int x;
int y;
int z;
int temp;
scanf("%d%d%d", &x, &y, &z);
if (x < y) {
temp = y;
y = x;
x = temp;
}
if (x < z) {
temp = z;
z = x;
x = temp;
}
if (y < z) {
temp = z;
z = y;
y = temp;
}
printf("%d %d %dn", x,y,z);
return 0;
}
(2)
int main() {
int m = 18;
int n = 24;
int r = 0;
while (r = m % n)//当r=0时,程序默认条件为假,跳出程序
{
m = n;
n = r;
}
printf("%dn", n);
return 0;
}



