One 、 The experiment purpose

1． Learn circular sentences for、while and do-while How to use statements .

2． Learn to use circular statements to implement various algorithms , For example, exhaustive method 、 Iterative method, etc .

Two 、 Experimental content

1  Find the minimum

This question requires the preparation of procedures , Find the smallest value in a given series of integers .

Input format ： The input first gives a positive integer in a line n, And then n It's an integer , Separated by spaces .. Output format ： In a row, press “min = minimum value ” Format output n The smallest of an integer .

2 Print the Jiu formula table

Here's a complete list of the bottom triangle and nine nines ：

1*1=1   

1*2=2   2*2=4   

1*3=3   2*3=6   3*3=9   

1*4=4   2*4=8   3*4=12  4*4=16  

1*5=5   2*5=10  3*5=15  4*5=20  5*5=25  

1*6=6   2*6=12  3*6=18  4*6=24  5*6=30  6*6=36  

1*7=7   2*7=14  3*7=21  4*7=28  5*7=35  6*7=42  7*7=49  

1*8=8   2*8=16  3*8=24  4*8=32  5*8=40  6*8=48  7*8=56  8*8=64  

1*9=9   2*9=18  3*9=27  4*9=36  5*9=45  6*9=54  7*9=63  8*9=72  9*9=81  

This question requires any given positive integer N, Output from 1*1 To N*N Part of the formula table

Output format ： Input gives a positive integer on a line N（1≤N≤9）

Input format ： Output the lower triangle N*N Part of the formula table , The number to the right of the equal sign is 4 position 、 Align left

3 Calculate factorial sum

For a given positive integer N, You need to calculate S=1!+2!+3!+...+N!.

Input format ： The input gives no more than 10 The positive integer N.

Output format ： Output in one line S Value .

4 Sum of integral segments

Given two integers A and B, Output from A To B All the integers of and the sum of these numbers .

Input format ： Enter... On one line 2 It's an integer A and B, among −100≤A≤B≤100, Separated by spaces .

Output format ： First, output sequentially from A To B All integers of , Every time 5 A line of numbers , Each number accounts for 5 Character width , Align right . Finally, in a line, press Sum = X Output the sum of all the numbers X.

5 Count the student's scores

This question requires writing a program to read in N A student's 100 point mark , Make statistics on the distribution of the scores on the five point system . The conversion rules from 100% to 5% ：

• Greater than or equal to 90 It is divided into A;

• Less than 90 Greater than or equal to 80 by B;

• Less than 80 Greater than or equal to 70 by C;

• Less than 70 Greater than or equal to 60 by D;

• Less than 60 by E.

Input format : Input gives a positive integer in the first line N（≤1000）, That is, the number of students ; Given in the second line N A student's 100 point mark , Separated by spaces . Output format : Output in one line A、B、C、D、E The distribution of the number of people corresponding to the five point system , Numbers are separated by spaces , There must be no extra space at the end of the line .

6 special a Summation of series of numbers (20 branch )

Given that neither is more than 9 The positive integer a and n, Ask to write a program a+aa+aaa++⋯+aa⋯a（n individual a） The sum of the .

Input format ：

The input is given in one line no more than 9 The positive integer a and n.

Output format ：

In a row, press “s = Corresponding sum ” Format output .

sample input ：

2 3

sample output ：

s = 246

7 Narcissistic number

The number of Narcissus refers to a N Positive integer （N≥3）, Of the numbers in each of its bits N The sum of the powers is equal to itself . for example ：153=1

​3

​​ +5

​3

​​ +3

​3

​​ . This question requires the preparation of procedures , Calculate all N Number of daffodils .

Input format :

Input gives a positive integer on a line N（3≤N≤7）.

Output format :

Output all... In ascending order N Number of daffodils , One line for each number .

sample input :

3

sample output :

153

370

371

407

8 Print hourglass (20 branch )

This question requires you to write a program to print the given symbol into the shape of an hourglass . For example, given 17 individual “*”, It is required to print in the following format

*****

 ***

  *

 ***

*****

So-called “ Hourglass shape ”, Refers to the output of odd symbols per line ; Align symbol centers ; Number difference between two adjacent lines of symbols 2; The number of symbols decreases from large to small to 1, Increase from small to large ; Equal number of first and last symbols .

Given arbitrary N Symbols , It doesn't have to be an hourglass . Require the printed hourglass to use as many symbols as possible .

Input format :

Type in on a line to give 1 A positive integer N（≤1000） And a symbol , Space between .

Output format :

First print out the largest hourglass shape made up of the given symbols , Finally, output the number of unused symbols in one line .

sample input :

19 *

sample output :

*****

 ***

  *

 ***

*****

2

3、 ... and 、 Screenshot of experimental source program and results

1 Find the minimum

#include <stdio.h>

int main()
{
    int n,i = 0,j,min;
    do
    {
        scanf("%d ",&n);
    }while(n <= 0);          // control n The scope of the 
    for(i = 0;i < n;i++)
    {
        scanf("%d ",&j);     // Input n It's an integer 
        if(i == 0)           
        {
            min = j;         // Make the first number equal to min
        }
        else
        {
            if(j < min)      // If j Than min Small then handle j The value is assigned to min
            {
                min = j;
            }
        }
        i++;
    }
    printf("min = %d",min);    // Output min 
    return 0;
}

2 Print the Jiu formula table

#include <stdio.h>

int main()
{
    int N,i,j;
    do
    {
        scanf("%d",&N);
    }while(N <= 0);        // control N The scope of the 
    for(i = 1;i <= N;i++)         // Control the number of lines printed 
    {
        for(j = 1;j <= i;j++)        // Control the of each line geshu
        {
            printf("%d*%d=%-4d",j,i,i*j);
        }
        printf("\n");       // Control line feed 
    }
    return 0;
}

3 Calculate factorial sum

#include <stdio.h>

int main()
{
    int N,sum = 0,S,i,j;
    do
    {
        scanf("%d",&N);    
    }while(N <= 0 || N > 10);       // control N The scope of the 
    for(i = 0;i < N;i++)           // The outer loop controls the number of cumulative multipliers 
    {
        for(j = 0;j <= i;j++)      // The inner loop controls the end point of each multiplication 
        {
            S = S * (j + 1);
        }
        sum = sum + S;             // Sum of multiplications 
        S = 1;                     // initialization S
    }
    printf("%d",sum);              // Output sum 
    return 0;
}

4 Sum of integral segments  

#include <stdio.h>

int main()
{
    int A,B,i,j = 1,sum = 0;
    do
    {
        scanf("%d %d",&A,&B);
    }while(A<-100||A>100 || B<-100||B>100||A>B);      // control A and B The scope of the 
    for(i = A;i <= B;i++,j++)           // from A To B All the numbers of 
    {
        printf("%5d",i);            // Output numbers and align them to the right within the width of five characters 
        sum = sum + i;              // Sum up 
        if(j % 5 == 0)
        printf("\n");               // Wrap every five numbers 
    }
    if((j - 1) % 5 != 0)            // If the last line is less than five, the output is wrapped 
    printf("\n");
    printf("Sum = %d",sum);
    return 0;
}

 5 Count the student's scores

#include <stdio.h>

int main()
{
    int N,i,j,a=0,b=0,c=0,d=0,e=0;     // Define initial values at all levels 
    do
    {
        scanf("%d",&N);
    }while(N > 1000 || N <= 0);            // control N The scope of the 
    for(i = 0;i < N;i++) 
    { 
        scanf("%d",&j);
        if(j >= 90)
        {
            a++;                  // Results are greater than 90 when a Add one 
        }
        else if(j >= 80 && j < 90)
        {
            b++;                  // Results are greater than 80 Less than 90 when b Add one 
        }
        else if(j >= 70 && j < 80)
        {
            c++;                  // Results are greater than 70 Less than 80 when b Add one 
        }
        else if(j >= 60 && j < 70)
        {
            d++;                  // Results are greater than 70 Less than 60 when b Add one 
        }
        else
        {
            e++;                  // The score is less than 60 when e Add one 
        }
    }
    printf("%d %d %d %d %d",a,b,c,d,e);
    return 0;
}

6 special a Summation of series of numbers

#include <stdio.h>

int main()
{
    int a,n,i;
    double s = 0.0,x = 0.0;
    do
    {
        scanf("%d %d",&a,&n);
    }while(a<=0 || a > 9 || n <= 0 || n > 9);       // control a and n The scope of the 
    for(i = 0;i < n;i++)
    {
        x = x*10 + a;                       // Calculate each aa……a Value 
        s = s + x;                         // Sum up 
    }
    printf("s = %.0lf",s);          // Output integer 
    return 0;
}

7 Narcissistic number

#include <stdio.h>
#include <math.h>
int main()
{
    int N,i,j,k,s=1,sum=0,e,a;
    scanf("%d",&N);
    for(i=pow(10,N-1);i<pow(10,N);i++)         // Find all N digit 
    {
        e=i;
        for(k=1;k<=N;k++)
        {
            a=i%10;           // Find the single digit 
            i=i/10;
            for(j=1;j<=N;j++)
            {
                s=s*a;             // count a Of N Power 
            }
            sum=sum+s;
            s=1;
        }
            if(sum==e)        // If you qualify 
        {
            printf("%d\n",e);
        }
        sum=0;
        i=e;
    }
    return 0;
}

8 Print hourglass

#include <stdio.h>
int main()
{
    int N,i,j=0,m,n,sum=0;
    char c;
    scanf("%d %c",&N,&c);
    for(i=1;1+2*(i*i-1)<=N;i++)
    {
        j++;
    }// Count the number of rows of the largest upper half of the hourglass 
    for(m=1;m<=j;m++)// Print the upper half of the hourglass 
    {
        for(n=1;n<=2*j-m;n++)
        {
            if(n<=m-1)// Each line prints a different number of spaces and characters 
                printf(" ");
            else
            {
                printf("%c",c);
                sum++;// Record how many characters are printed 
            }
        }
        printf("\n");
    }
    for(m=1;m<j;m++)// Print the bottom half of the hourglass 
    {
        for(n=1;n<=j+m;n++)
        {
            if(n<=j-1-m)// Print different spaces and characters on each line 
                printf(" ");
            else
            {
                printf("%c",c);
                sum++;// Record how many characters are printed 
            }
        }
        printf("\n");// Line break 
    }
        printf("%d",N-sum);// Output the number of characters remaining 
    return 0;
}

Four 、 Analysis and thinking of experiment

1、 Find the minimum

   The first number is equal to the minimum , Then compare the following numbers through a loop , If it is less than the minimum value, it is assigned to the minimum value .

2 Print the Jiu formula table

   Double loops are used to control rows and columns respectively .

3 Calculate factorial sum

   First use the loop to carry out factorial operation , Then sum it .

4 Sum of integral segments

   Print the number between the two numbers , A number occupies five digits and records the number of prints , Whenever the number of prints is 5 When it's a multiple of , Wrap once , When the total number of printing times is a multiple of five, there is no line break at last .

5 Count the student's scores

   According to the students' grades, compare with the conditions through cycles , If the conditions are met, the corresponding number shall be increased by one .

6 special a Summation of series of numbers

   First use the loop to find aa...a Value , And then sum up

7 Narcissistic number

Count out every digit , Use the cycle to find their N Power , And then sum up , Judge whether the conditions are met .

8 Print hourglass

   First calculate the maximum number of lines that can print the upper half of the hourglass , Then print the upper half of the hourglass according to the quantitative relationship , Then print the hourglass of the off-duty part according to the quantity relationship .

hypothesis N=5, The maximum number of lines in the upper half is j=3

Row number    Space    character  sum   So the number of lines is sum The relationship is sum=N+1- Row number

  1      0       5    5         The relationship between the number of spaces and the number of lines is that the number of spaces is equal to the number of lines -1

  2      1       3    4

3      2       1    3     

Row number    Space    character  sum   So the number of lines is sum The relationship is sum= Row number +j

 1       1       3  4       The relationship between the number of spaces and the number of lines is that the number of spaces is equal to j-1- Row number

 2       0       5  5       As the number of spaces increases, add the number of lines , Gradually reduce the number of lines