"Introduction to the algorithm contest" on the computer practice--The first chapter _ Algorithm competition

Exercise 1-1 average (average)

Enter 3 integers, output their average value, and keep 3 decimal places.

#include <stdio.h>

int main ()
{
    int a,b,c;
    Double average;
    
    scanf ("%d%d%d", &a,&b,&c);
    Average = (a+b+c)/3.0;
    printf ("%.3lf\n", average);
    
    return 0;
}

Exercise 1-2 temperature (temperature)

Enter Fahrenheit temperature F, output corresponding to Celsius temperature C, retain 3 decimal places. Tip: C=5 (f-32)/9.

#include <stdio.h>

int main ()
{
    double f,c;
    
    scanf ("%lf", &f);
    c = 5* (f-32)/9;
    printf ("%.3lf\n", c);
    
    return 0;
}

Exercise 1-3 continuous and (sum)

Input positive integer n, output 1+2+ The value of the +n. Tip: The goal is to solve the problem, not to practice programming.

Analysis: 1+2+ +n = (1+n) *N/2

#include <stdio.h>

int main ()
{
    int n,sum;
    
    scanf ("%d", &n);
    sum = (1+n) *n/2;
    printf ("%d\n", sum);
    
    return 0;
}

Exercise 1-4 sine and cosine (sincos)

Input positive integer n (n<360), output the sine, cosine function value of n degrees. Tip: Use mathematical functions.

Analysis: The angle must be converted to radians. Angle 180 degree corresponds to Radian 1pi

#include <stdio.h>
#include <math.h>

int main ()
{
    double n/* angle * * *,
	       x/* radians.
	Const double PI = 4.0*atan (1.0);
    
    scanf ("%lf", &n);
    x = N/180*pi;
    printf ("%lf\n", Sin (x));
    printf ("%lf\n", cos (x));
    
    return 0;
}

Exercise 1-5 distance (distance)

Enter 4 floating-point number x1,y1,x2,y2 and output the distance of the midpoint (x1,y1) point (x2,y2) of the planar coordinate system.

Analysis: Plane distance between two points: D=sqrt ((x2-x1) * (x2-x1) + (y2-y1) * (y2-y1))

sqrt () Calculate the square root of arithmetic

#include <stdio.h>
#include <math.h>

int main ()
{
    double x1,y1,x2,y2,d;
    
    scanf ("%lf%lf%lf%lf", &x1,&y1,&x2,&y2);
    D = sqrt ((x2-x1) * (x2-x1) + (y2-y1) * (y2-y1));
    printf ("%lf\n", d);
    
    return 0;
}

Exercise 1-6 even (odd)

Enter an integer to determine if it is an even number. If it is, the output is "yes", otherwise "no" is output. Hint: can be judged by a variety of methods.

Analysis: In addition to% 2, today we learned a new method: if (n&1), with 1 bitwise AND, the result is 1 is odd, 0 is even

#include <stdio.h>

int main ()
{
    int n;
    
    scanf ("%d", &n);
    if (n&1) printf ("No"),/* is odd/
    else printf ("yes"),/* is even/return 
    
    0;
}

Exercise 1-7 Discount (discount)

A clothes 95 yuan, if the consumption of 300 yuan, can play 85 percent. Enter the number of pieces of clothing purchased, output the amount to be paid (in units: yuan), retain two decimal places.

#include <stdio.h>

int main ()
{
    int n;
    Double money;
    
	scanf ("%d", &n);
	Money = 95*n;
	if (money>=300)
	    money*=0.85;
	printf ("%.2lf\n", money);
	
    return 0;
}

Exercise 1-8 Absolute Value (ABS)

Enter a floating-point number, output its absolute value, and retain two decimal places.

#include <stdio.h>

int main ()
{
    double n;
    
	scanf ("%lf", &n);
	if (n>0) 
	    printf ("%.2lf\n", n);
	else 
	    printf ("%.2lf\n",-N);
	
    return 0;
}

Exercise 1-9 triangle (triangle)

Enter a triangular three-side length value (all positive integers) to determine whether it can be three sides long for a right triangle. If so, the output is "yes" and if not, the "no" is output. If the triangle cannot be formed at all, the "not a triangle" is output.

Analysis: First to determine whether the triangle can be formed, and further determine whether to constitute a right-angled triangle.

#include <stdio.h>

int main ()
{
    int a,b,c,t;
    
    scanf ("%d%d%d", &a,&b,&c);
    if (a>c)
        {t=a;a=c;c=t;}
    if (b>c)
        {t=b;b=c;c=t;} /* These 2 if statements ensure that C is the longest edge */ 
        
    if (a+b<=c)
        printf ("Not a triangle\n");
    else
        if (a*a+b*b = = c*c)
            printf ("yes\n");
        else
            printf ("no\n");
    
    return 0;
}

Exercise 1-10 years (year)

Enter the year to determine if it is a leap years. If it is, the output is "yes", otherwise "no" is output. Hint: It is not enough to simply judge the remainder divided by 4.

#include <stdio.h>

int main ()
{
    int y;
    
    scanf ("%d", &y);
    if (y%4==0 && y%100!=0 | | | y%400==0)
        printf ("yes\n");
    else
        printf ("no\n");
    
    return 0;
}