#include <stdio.h>
int main(void)
{
int num;
scanf("%d", &num);
if (num > 0) {
printf("positive\n");
} else if (num == 0) {
printf("zero\n");
} else {
printf("negative\n");
}
return 0;
}
#include <stdio.h>
#include <math.h>
int main(void)
{
double num;
scanf("%lf", &num);
num = fabs(num);/* 调用了 fabs() 这一个求绝对值的函数*/
printf("%.2lf\n", num);
return 0;
}
#include <stdio.h>
int main(void)
{
int num;
scanf("%d", &num);
if (num % 2 == 0)
printf("even\n");
else
printf("odd\n");
return 0;
}
#include<stdio.h>
int main(void)
{
char c;
scanf("%c", &c);
if(c % 2 == 0)
printf("NO\n");
else
printf("YES\n");
return 0;
}
#include <stdio.h>
int main(void)
{
/* 注意一下 x 和 y 的范围 */
long long x, y;
scanf("%lld%lld", &x, &y);
if (x > y) {
printf(">\n");
} else if (x == y) {
printf("=\n");
} else {
printf("<\n");
}
return 0;
}
#include <stdio.h>
int main(void)
{
int num;
scanf("%d", &num);
if (num >= 10 && num <= 99)
printf("1\n");
else
printf("0\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int num_1, num_2;
scanf("%d%d", &num_1, &num_2);
if (num_1 >= 10 || num_2 >= 20)
printf("1\n");
else
printf("0\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int num;
scanf("%d", &num);
if ((num % 3 == 0) && (num % 5 == 0))
printf("YES\n");
else
printf("NO\n");
return 0;
}
#include <stdio.h>
#include <stdbool.h>
int main(void)
{
bool flag = false;/* 设定一个标志位 */
int num;
scanf("%d", &num);
if (num % 3 == 0) {
printf("3 ");
flag = true;
}
if (num % 5 == 0) {
printf("5 ");
flag = true;
}
if (num % 7 == 0) {
printf("7 ");
flag = true;
}
if (flag == false)
printf("n\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int counter = 0; /* 记录不及格科目的数量 */
int score_1, score_2;
scanf("%d%d", &score_1, &score_2);
if (score_1 < 60)
counter++;
if (score_2 < 60)
counter++;
if (counter == 1)
printf("1\n");
else
printf("0\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int day;
scanf("%d", &day);
if (day == 1 || day == 3 || day == 5)
printf("NO\n");
else
printf("YES\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int distance;
scanf("%d", &distance);
if (distance < 100) { /* 用数学方法可以求得临界距离为100 */
printf("Walk\n");
} else if (distance == 100) {
printf("All\n");
} else {
printf("Bike\n");
}
return 0;
}
#include <stdio.h>
int main(void)
{
double x;
scanf("%lf", &x);
double y = 0.0;
if (x >= 0 && x < 5) {
y = -x + 2.5;
} else if (x >= 5 && x < 10) {
y = 2 - 1.5 * (x - 3) * (x - 3);
} else {
y = x / 2 - 1.5;
}
printf("%.3lf\n", y);
return 0;
}
#include <stdio.h>
int main(void)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if (a > b) {
int temp = a;
a = b;
b = temp;
}
if (b > c) {
int temp = b;
b = c;
c = temp;
}
printf("%d\n", c); /* 经过两轮交换,c一定是最大的 */
return 0;
}
#include <stdio.h>
int main(void)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if ((a + b > c) && (a - b < c))
printf("yes\n");
else
printf("no\n");
return 0;
}
#include <stdio.h>
int main(void)
{
int year;
scanf("%d", &year);
/* 闰年是能被400整除,或者是能被4整除但是不能被100整除的年份 */
if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0))
printf("Y\n");
else
printf("N\n");
return 0;
}
#include<stdio.h>
int main(void)
{
int x, y;
scanf("%d%d", &x, &y);
if(-1 <= x && x <= 1 && -1 <= y && y <= 1)
printf("yes");
else
printf("no");
return 0;
}
#include <stdio.h>
int main(void)
{
int a, b, c;
char ch;
scanf("%d%d %c", &a, &b, &ch);/* 在%c前加一个空格,不然程序会出问题 */
switch (ch)
{
case '+':
{
c = a + b;
printf("%d\n", c);
break;
}
case '-':
{
c = a - b;
printf("%d\n", c);
break;
}
case '*':
{
c = a * b;
printf("%d\n", c);
break;
}
case '/':
{
if (b == 0) {
printf("Divided by zero!\n");
} else {
c = a / b;
printf("%d\n", c);
}
break;
}
default:
{
printf("Invalid operator!\n");
break;
}
}
return 0;
}
#include<stdio.h>
#include<math.h>
int main(void)
{
float a, b, c;
scanf("%f%f%f", &a, &b, &c);
if(b * b == 4 * a * c)
{
if(b == 0)
printf("x1=x2=%.5f\n", b / (2 * a));
else
printf("x1=x2=%.5f\n", (-b) / (2 * a));
}
else if(b * b > 4 * a * c)
printf("x1=%.5f;x2=%.5f\n", (-b + sqrtf((b * b - 4 * a *c))) / (2 * a), (-b - sqrtf(b * b - 4 * a * c)) / (2 * a));
else
{
if(b == 0)
printf("x1=%.5f+%.5fi;x2=%.5f-%.5fi\n", b / (2 * a), sqrtf(4 * a * c - b * b) / (2 * a), b / (2 * a), sqrtf(4 * a * c - b * b) / (2 * a));
else
printf("x1=%.5f+%.5fi;x2=%.5f-%.5fi\n", -b / (2 * a), sqrtf(4 * a * c - b * b) / (2 * a), -b / (2 * a), sqrtf(4 * a * c - b * b) / (2 * a));
}
return 0;
}
#include <stdio.h>
int main(void)
{
int n, x, y;
scanf("%d%d%d", &n, &x, &y);
if (y >= n * x) /* 苹果吃完了 */
printf("0\n");
else if (y % x == 0) /* 苹果没有吃完 */
printf("%d\n", n - y / x);
else
printf("%d\n", n - y / x - 1);
return 0;
}
博客仅供参考
OpenJudge - NOI - 1.4编程基础之逻辑表达式与条件分支(C语言 全部题解)
原文:https://www.cnblogs.com/liuzhaotong/p/12193975.html