"Algorithmic Competition Primer Classic" 1.5.6 On-machine practice

Exercise 1-1 average (average)

1#include <stdio.h>2 intMain ()3 {4     inta,b,c;5scanf"%d%d%d",&a,&b,&c);6printf"%0.3LF", (A+B+C)/3.0);7     return 0;8}

The focus is on format output, type conversion ...

But in

printf ("%0.3lf", (A+B+C)/3), when input 1 1 2 is output 0, this I really do not understand, int type is not to get 1 and then double ...

Exercise 1-2 temperature (temperature)

1#include <stdio.h>2 intMain ()3 {4     floatf,c;5scanf"%f",&f);6printf"%0.3f",5* (f +)/9.0);7     return 0;8}

This in

scanf ("%f",&f); Change%f to%d, input 32 also error, result output:-17.778. Normal is not supposed to be 0 ... Although there is an int and float conversion, but it does not feel ah ...

I searched the web http://tieba.baidu.com/p/88194315

float a=3; printf ("a=%d\n", a) results a=0 ah? Ask for advice because you define a as a Folat type, but your output is int type. The output computer automatically converts the value of the float type to int, which is to assign the latter 16 bits of the float to the int number. In this question, the 16 bits of the float number are all zeros. So the output is 0. Or a forced type conversion. Or with the%f type output??

Feel the sense of that.

Exercise 1-3 continuous and (sum)

1#include <stdio.h>2 intMain ()3 {4     intN;5scanf"%d",&n);6printf"%d", N (n+1)/2);7     return 0;8}

Exercise 1-4 sine and cosine (sincos)

1#include <stdio.h>2#include <math.h>3 intMain ()4 {5     Const floatPi=4.0*atan (1.0);6     intN;7scanf"%d",&n);8printf"%LF%LF", Sin (n*pi/ the), cos (n*pi/ the) );9     return 0;Ten}

The key is math ...

Exercise 1-5 distance (distance)

1#include <stdio.h>2#include <math.h>3 intMain ()4 {5     floatx1, x2, y1, y2;6     floats;7scanf"%f%f%f%f",&x1,&y1,&x2,&y2);8S=sqrt ((x1-x2) * (X1-X2) + (y1-y2) * (y1-y2));9printf"%f", s);Ten     return 0; One}

"Algorithmic Competition Primer Classic" 1.5.6 On-machine practice