1 Charactor arrays

Know the difference between Char arrays and strings.
Define the common string functions which are included in string.h

// write your own common string functions
#include <stdio.h>
#include <string.h>

int strlen_def(char *str){
    int i=0;
    while(str[i]!='\0') i++;
    return(i);
}

void strcat_def(char *a1, char *a2){
    int i, len_a1 = strlen_def(a1);
    int len_a2 = strlen_def(a2);
    //a1[len_a1++] = ' ';
    for (i = 0; i <= len_a2; i++){
        a1[len_a1 + i] = a2[i];
    }
    a1[len_a1+i] = '\0';
}

void strcpy_def(char *a1, char *a2){
    int i, len_a1 = strlen_def(a1);
    int len_a2 = strlen_def(a2);
    if(len_a1<len_a2+1)
        printf("The length of the first string must be longer than the second.");
    else{
        for (i = 0; i < len_a2; i++){
            a1[i] = a2[i];
        }
        a1[i] = '\0';
    }
}

int strcmp_def(char *a1, char *a2){
    int i=0, len_a1 = strlen_def(a1);
    int len_a2 = strlen_def(a2), minlen = (len_a1>len_a2)?len_a2:len_a1;
    while(i<minlen){
        if(a1[i]<a2[i]) return -1;
        else if(a1[i]>a2[i]) return 1;
        else i++;
    }
    return 0;
}

void strlwr_def(char *a1){
    int i=0;
    while(a1[i]!='\0'){
        if(a1[i]>='A'&& a1[i]<='Z') a1[i]+=32;
        i++;
    }
}

void strupr_def(char *a1){
    int i=0;
    while(a1[i]!='\0'){
        if(a1[i]>='a' && a1[i]<='z') a1[i]-=32;
        i++;
    }
}

void gets_def(char *a1){
    int i=0;
    char c;
    scanf("%c",&c);
    while(c!='\n'){
        a1[i++] = c;
        scanf("%c",&c);
    }
    a1[i] ='\0';
}

void puts_def(char *a1){
    int i=0;
    while(a1[i]!='\0')
        printf("%c",a1[i++]);
}

int main()
{
    char str1[100] = "I am a student.";
    char str2[] = "I am Chinese.";
    strcpy_def(str1,str2);
    //strcat_def(str1,str2);
    //strlwr_def(str1);
    //gets_def(str1);
    printf("%s\n",str1);
    //puts_def("I am a student.\nI am Chinese.");

    //printf("%d\t%d\n",strlen_def(str1),strlen(str1));
    //printf("%d\n",strcmp_def(str1,str2));

    return 0;
}

Write a function that can input continuously many sentences from I/O system.

#include <stdio.h>
#include <string.h>

void inputstr(char *p){
    int i = 0;
    char temp[1000];
    while (i<100){
        gets(temp);
        
        if (!(strcmp(temp, "The end!") && strcmp(temp, "the end") 
            && strcmp(temp, "the end!") && strcmp(temp, "end")))
            break;
        strcat(p, temp);
        strcat(p, " ");
        i++;
    }
}

int main()
{
    char str1[10000]={'\0'};
    inputstr(str1);
    printf("%s",str1);  

    fflush(stdin);
    getchar();
    return 0;
}

2 Functions

The syntax of functions
Declaration of functions following the #include <>

#include <stdio.h>
int max4(int a,int b,int c,int d);
int max2(int a,int b);

int main()
{
    int a,b,c,d,max;
    printf("Please enter 4 interger numbers:");
    scanf("%d %d %d %d",&a,&b,&c,&d);
    max=max4(a,b,c,d);
    printf("max=%d \n",max);
    return 0;
}

int max4(int a,int b,int c,int d)
{
    int m;
    m=max2(a,b);
    m=max2(m,c);
    m=max2(m,d);
    return(m);
}

int max2(int a,int b)
{  if(a>=b)
        return a;
    else
        return b;
}

3 Sugar functions

3.1 There are 10 students with 5 courses.

Calculate the average score of each student.
Calculate the average score of each course.
Find the highest score and the corresponding student and course
Calculate the varance of the student’s average scores. \[\begin{align} \sigma = \frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2 \end{align}\] where \(x_i\) is the student’s average score.

#include <stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 10
#define M 5
float score[N][M];
float a_student[N],a_course[M];
int r,c;

int main()
{
    int i,j,seed = 1;
    float h;
    float var();
    float highest();
    void input_student(int seed);
    void aver_student();
    void aver_course();

    input_student(seed);
    aver_student();
    aver_course();
    printf("\n  NO.      course1    course2   course3   course4   course5   aver\n");
    for(i=0;i<N;i++){
        printf("\n NO %2d ",i+1);
        for(j=0;j<M;j++)
            printf("%10.2f",score[i][j]);
        printf("%10.2f\n",a_student[i]);
    }
    printf("\naverage:");
    for (j=0;j<M;j++)
        printf("%8.2f",a_course[j]);
    printf("\n");
    h=highest();
    printf("highest:%7.2f   NO. %2d   coursese %2d\n",h,r,c);
    printf("variance %8.2f\n",var());
    return 0;
}

void input_student(int seed)
{
    int i,j;
    printf("\nthe scores of students:\n");
    srand(seed);
    for(i=0;i<N;i++){
        for (j=0;j<M;j++)
          score[i][j] = 60+rand()%40;
    }
}


void aver_student()
{
    int i,j;
    float s;
    for(i=0;i<N;i++){
        for (j=0,s=0;j<M;j++)
            s+=score[i][j];
        a_student[i]=s/5.0;
    }
}

void aver_course()
{
    int i,j;
    float s;
    for(j=0;j<M;j++){
        s=0;
        for(i=0;i<N;i++)
            s+=score[i][j];
        a_course[j]=s/(float)N;
    }
}

float highest()
{
    int i,j;
    float high = score[0][0];
    for(i=0;i<N;i++)
        for(j=0;j<M;j++){
            if(score[i][j]>high){
                high=score[i][j];
                 r=i+1;
                 c=j+1;
            }
        }
    return(high);
}

float var()
{
    int i;
    float sumx = 0.0, sumxn = 0.0;
    for(i=0;i<N;i++){
        sumx+=a_student[i]*a_student[i];
        sumxn+=a_student[i];
    }
    return(sumx/N-(sumxn/N)*(sumxn/N));
}

#include <stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 10
#define M 5

int main()
{
    int i,j,seed = 1;
    int r,c;
    float h, score[N][M],a_student[N],a_course[M];
    float var(float *a_student);
    float highest(float score[N][M], int *r, int *c);
    void input_student(float score[N][M], int seed);
    void aver_student(float score[N][M], float *a_student);
    void aver_course(float score[N][M], float *a_course);

    input_student(score, seed);
    aver_student(score,a_student);
    aver_course(score, a_course);
    printf("\n  NO.      course1    course2   course3   course4   course5   aver\n");
    for(i=0;i<N;i++){
        printf("\n NO %2d ",i+1);
        for(j=0;j<M;j++)
            printf("%10.2f",score[i][j]);
        printf("%10.2f\n",a_student[i]);
    }
    printf("\naverage:");
    for (j=0;j<M;j++)
        printf("%10.2f",a_course[j]);
    printf("\n");
    h=highest(score, &r, &c);
    printf("highest:%7.2f   NO. %2d   coursese %2d\n",h,r,c);
    printf("variance %8.2f\n",var(a_student));
    return 0;
}

void input_student(float score[N][M],int seed)
{
    int i,j;
    printf("\nthe scores of students:\n");
    srand(seed);
    for(i=0;i<N;i++){
        for (j=0;j<M;j++)
          score[i][j] = 60+rand()%40;
    }
}


void aver_student(float score[N][M], float *a_student)
{
    int i,j;
    float s;
    for(i=0;i<N;i++){
        for (j=0,s=0;j<M;j++)
            s+=score[i][j];
        a_student[i]=s/M;
    }
}

void aver_course(float score[N][M], float *a_course)
{
    int i,j;
    float s;
    for(j=0;j<M;j++){
        s=0;
        for(i=0;i<N;i++)
            s+=score[i][j];
        a_course[j]=s/N;
    }
}

float highest(float score[N][M], int *r, int *c)
{
    int i,j;
    float high = score[0][0];
    for(i=0;i<N;i++)
        for(j=0;j<M;j++){
            if(score[i][j]>high){
                high=score[i][j];
                 *r=i+1;
                 *c=j+1;
            }
        }
    return(high);
}

float var(float *a_student)
{
    int i;
    float sumx = 0.0, sumxn = 0.0;
    for(i=0;i<N;i++){
        sumx+=a_student[i]*a_student[i];
        sumxn+=a_student[i];
    }
    return(sumx/N-(sumxn/N)*(sumxn/N));
}

3.2 Calculate \(m\) to power \(n\)

that is, \(m^n\), where both \(m\) and \(n\) are positive integers.

/*pow(M,N)*/
#include<stdio.h>
#define M 2
#define N 100
#define L 1000
void fun(int arr[]){
    int i,a,b=0;
    for(i=L-1;i>=0;i--){
        a=arr[i]*M+b;
        arr[i]=a%10;
        b=a/10;
    }
}

int main(){
    int i,j,a[L]={0};
    a[L-1]=M;
    for(i=1;i<N;i++) fun(a);
    for(i=0;i<L;i++){
        if(a[i]>0){
            j=i;
            break;
        }
    }
    printf("pow(%d,%d)=",M,N);
    for(j=i;j<L;j++)
        printf("%d",a[j]);
    printf("\ni=%d\n",i);
    return 0;
}

4 Problem in the previous week

Calculate the first ten pairs \((p,q)\) satisfying \(p^2-8q^2=1\). More details about Pell’s equation can be found at https://en.wikipedia.org/wiki/Pell%27s_equation.
There are three simple methods to arrive at this goal.

// Calculate the first 10 pairs (p,q) of Pell's equation.
#include<stdio.h>
void pf8iter(int max_iter){
    int p=3,q=1,count=1,tmp1;
    printf("(p,q) = (%d,%d)\n",p,q);
    while(count<max_iter){
        tmp1 = 3*p+8*q;
        q = 3*q+p;
        p=tmp1;
        count++;
        printf("(p,q) = (%d,%d)\n",p,q);
    }
}

int pf8int1(int max_iter){
    int p=1,q=1,count=0,times=0,f0=p*p-8*q*q-1;
    while(count<max_iter){
        times++;
        if(f0<0){
            f0=f0+4*(p+1);
            p+=2;
        }
        else if(f0>0){
            f0=f0-8*(2*q+1);
            q++;
        }
        else{
            printf("(p,q) = (%d,%d)\n",p,q);
            count++;
            f0=4*(p+1)-8*(2*q+1);
            p+=2;
            q++;
        }
    }
    printf("times = %d\n",times);
    return times;
}

int ifsquare(int x) {
    // if x is a square number, return its square root; else return 0;
    int i = (int) sqrt((float) x);
    int isquare = i * i;
    while (isquare < x) { isquare = isquare + 2 * i + 1; i++; }
    if (isquare == x) return i;
    else return 0;
}

int pf8new(int max_iter) {
    int t, p, q, tsquare, i, indicator,times=0;
    i = 0, t = 1, tsquare = t * t;
    while (i < max_iter) {
        times++;
        // tsquare = t*t; test if (tsquare+1)/2 is a square number;
        indicator = ifsquare( (tsquare + 1) / 2 );
        if (indicator)
        {
            p = tsquare * 2 + 1;
            q = t * indicator;
            printf("(p,q) = (%d,%d)\n",p,q);
            i++;
        }
        else {
            // if we define x = tsquare - 1, then x + 1 is a square number.
            // test if x/2 (which equal to (tsquare - 1)/2) is a square number;
            indicator = ifsquare( (tsquare - 1) / 2 );
            if (indicator)
            {
                p = tsquare * 2 - 1; // p = x * 2 + 1;
                q = t * indicator;
                printf("(p,q) = (%d,%d)\n",p,q);
                i++;
            }
        }
        tsquare = tsquare + t * 4 + 4; // set tsquare = (t + 2)*(t + 2);
        t = t + 2;
    }
    printf("times = %d\n",times);
    return times;
}

int main(){    
    pf8int1(10);
    pf8new(10);
    pf8iter(10);
    
    return 0;
}

5 Problems

A magic square is a square grid (where n is the number of cells on each side) filled with distinct positive integers in the range. such that each cell contains a different integer and the sum of the integers in each row, column and diagonal is equal, where the sum equals \(n(n^2+1)/2\).
find the first fifty pairs \((p,q)\) satisfying \(p^2-8q^2=1\). More details about Pell’s equation can be found at https://en.wikipedia.org/wiki/Pell%27s_equation.

Define functions

Xu Liu

1 Charactor arrays

2 Functions

3 Sugar functions

3.1 There are 10 students with 5 courses.

3.2 Calculate \(m\) to power \(n\)

4 Problem in the previous week

5 Problems