C语言零基础入门习题（七）谁在说谎前言一、There is only one thief among the four suspects A,B,C,D. and each of them can explain only once.二、Write a program that reads integers, finds the largest of them, and counts its occurrences. Assume that the input ends with number

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概述

前言

C语言是大多数小白走上程序员道路的第一步，在了解基础语法后，你就可以来尝试解决以下的题目。放心，本系列的文章都对新手非常友好。

Tips：题目是英文的，但我相信你肯定能看懂

一、There is only one thief among the four suspects A,B,C,D. and each of them can explain only once.

A said: it's not me.

B said: it's C.

C said: it's B.

D: C is talking nonsense.

Three of the four are known to have told the truth. Please program to find out who is the thief, output the corresponding capital letters.

#include <stdio.h>

int main()
{
    char i;
    for (i='A';i<='D';i++)
    {
        if ((i!='A')+(i=='C')+(i=='B')+(i!='B')==3)
            printf("%c",i);
    }
    return 0;
}

二、Write a program that reads integers, finds the largest of them, and counts its occurrences. Assume that the input ends with number 0. Suppose that you entered 3 5 2 5 5 5 0; the program finds that the largest is 5 and the occurrence count for 5 is 4. show “no input” if input 0 only.

Here are sample runs:

<output1>

3 5 2 5 5 5 0 <enter icon>

largest number:5 occurrences count:4

<end output1>

<output2>

-3 -5 -2 -4 -2 -6 0 <enter icon>

largest number:-2 occurrences count:2

<end output2>

<output3>

0 <enter icon>

No input

<end output3>

#include <stdio.h>
#include <limits.h>

int main()
{
    int a,b=INT_MIN,i=1;
    while (scanf("%d",&a) && a!=0)
    {
        if (b==a)
            i++;
        if (a>b)
            b=a;
    }
    if (b==0&&a==0)
        printf ("No input");
    else
        printf("largest number:%d occurrences count:%d",b,i);
    return 0;
}

三、Input a positive integer and output all the prime factors of the number with ascending order. You can stop it when input a negative number such as -1.

Here are sample runs:

<output>

120<enter icon>

2 2 2 3 5

6<enter icon>

2 3

32767<enter icon>

7 31 151

-3<enter icon>

Done

<end output>

#include <stdio.h>

int main()
{
    int a,i;
    while (scanf("%d",&a) && a>=0)
    {
        for (i=2;i<=a;i++)
        {
            if (a%i==0)
            {
                printf ("%d ",i);
                a=a/i;
                i=1;
            }
        }
        putchar('n');
    }
    printf("Done");
    return 0;
}

四、Armstrong number is a number that is equal to the sum of digits raise to the power total number of digits in the number. Some Armstrong numbers are: 153, 370, 407, 1634, 8208, etc.

371 = 3^3 + 7^3 + 1^3 (27 + 343 +1)

8208 = 8^4 + 2^4 +0^4 + 8^4

please input a positive number N (3≤N≤7), and output all Armstrong number with N digits.

Here are sample runs:

<output1>

3<enter icon>

153 370 371 407

<end output1>

<output2>

7<enter icon>

1741725 4210818 9800817 9926315

<end output2>

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

int main()
{
    int a,b,c,d,i,t;
    scanf ("%d",&a);
    for (i=pow(10,a-1);i<pow(10,a);i++)
    {
        c=i,d=i;
        for (t=a;t>0;t--)
        {
            b=pow(d%10,a);
            c-=b;
            d=d/10;
        }
        if (c==0)
            printf ("%d ",i);
    }
    return 0;
}

五、 (Reduce fractions) Write a program that prompts the user to enter the numerator and denominator of a fraction. The program determines whether the number is a proper fraction or an improper fraction. If it is a proper fraction, display the number. If not, reduce it to a mixed irreducible fraction or to an integer. The program will be ended with the input for denominator is 0.

Here are sample runs:

<output>

Enter a numerator: 45 <enter icon>

Enter a denominator: 46 <enter icon>

45 / 46 is a proper fraction

Enter a numerator: 45 <enter icon>

Enter a denominator: 15 <enter icon>

45 / 15 is an improper fraction and it can be reduced to 3

Enter a numerator: 45 <enter icon>

Enter a denominator: 25 <enter icon>

45 / 25 is an improper fraction and its mixed fraction is 1 + 4 / 5

Enter a numerator: 45 <enter icon>

Enter a denominator: 0 <enter icon>

Done

<output>

#include <stdio.h>

int main()
{
     int a,b,c,d,e,i;
     while (1)
     {
        printf("Enter a numerator: ");
        scanf("%d",&a);
        printf("Enter a denominator:");
        scanf("%d",&b);
        if (b==0) break;
        else
        {
            c=a-a/b*b;
        if (a<b)
            printf("%d / %d is a proper fractionn",a,b);
        else if (c==0)
            printf("%d / %d is an improper fraction and it can be reduced to %dn",a,b,a/b);
        else
        {
            i=c<b? b: c;
            for (;i>1;i--)
            {
                if (c%i==0&&b%i==0)
                    d=c/i,e=b/i;
            }
            printf("%d / %d is an improper fraction and its mixed fraction is %d + %d / %dn",a,b,a/b,d,e);
        }
        }
     }
     printf ("Done");
    return 0;
}