/*rain.c 针对若干年的降水量数据，计算年降水总量、年降水平均量，以及月降水平均量*/#include <stdio.h>#define MONTHS 12#define YEARS 5int main (void){    //把数组初始化为2000到2004年的降水量数据    const float rain[YEARS][MONTHS] = //const声明和初始化数组可保护数据    {        {4.3, 4.3, 4.3, 3.0, 2.0, 1.2, 0.2, 0.2, 0.4, 2.4, 3.5, 6.6},        {8.5, 8.2, 1.2, 1.6, 2.4, 0.0, 5.2, 0.9, 0.3, 0.9, 1.4, 7.3},        {9.1, 8.5, 6.7, 4.3, 2.1, 0.8, 0.2, 0.2, 1.1, 2.3, 6.1, 8.4},        {7.2, 9.9, 8.4, 3.3, 1.2, 0.8, 0.4, 0.0, 0.6, 1.7, 4.3, 6.2},        {7.6, 5.6, 3.8, 2.8, 3.8, 0.2, 0.0, 0.0, 0.0, 1.3, 2.6, 5.2}    };    int years,months;    float totyear, totmonth;    printf("YEAR  RAIN:\n");    for(years=0; years<YEARS; years++)    {        for(months=0,totyear=0; months<MONTHS; months++)            totyear += *(*(rain+years)+months);　　//此处应题目要求使用指针形式，同样可以用数组形式        printf("%4d %4.2f\n",2000+years,totyear);    }    printf("\nmonth average:\n");    printf("JUN  FEB  MAT  APR  MAY  JUN  JLY  AUG  SPT  OCT  NOV  DEC:\n");    for(months=0; months<MONTHS; months++)    {        for(years=0,totmonth=0; years<YEARS; years++)            totmonth += *(*(rain+years)+months);　　//可用数组形式        printf("%-4.1f ",totmonth/5);    }    printf("\nDone!\n");    return 0;}

/*一维数组函数调用*/#include <stdio.h>void copy_arr (double [], double [], int);void copy_ptr (double *, double *, int);int main (void){    double source[] = {1.1, 2.2, 3.3, 4.4, 5.5};    double target1[5] = {0};    //初始化一个元素全为0的数组可以这样赋值    double target2[5] = {0};    printf ("Before operation:\n");    printf ("source: \t%g\t%g\t%g\t%g\t%g\n",         source[0], source[1], source[2], source[3], source[4]);    printf ("target1: \t%g\t%g\t%g\t%g\t%g\n",         target1[0], target1[1], target1[2], target1[3], target1[4]);    printf ("target2: \t%g\t%g\t%g\t%g\t%g\n",         target2[0], target2[1], target2[2], target2[3], target2[4]);    copy_arr (source, target1, 5);    copy_ptr (source, target2, 5);    printf ("\n\nAfter operation:\n");    printf ("source: \t%g\t%g\t%g\t%g\t%g\n",         source[0], source[1], source[2], source[3], source[4]);    printf ("target1: \t%g\t%g\t%g\t%g\t%g\n",         target1[0], target1[1], target1[2], target1[3], target1[4]);    printf ("target2: \t%g\t%g\t%g\t%g\t%g\n",         target2[0], target2[1], target2[2], target2[3], target2[4]);        return 0;}void copy_arr (double a1[], double a2[], int n){    int i;    for (i = 0; i < n; i++)        a2[i] = a1[i];　　//简单的赋值就可以实现复制数组}void copy_ptr (double *p1, double *p2, int n){    int i;    for (i = 0; i < n; i++)        *(p2 + i) = *(p1 + i);　　//同样，赋值即可实现复制数组}/*******************************************PS: 有关%g输出的补充说明************************************************///1.%g用于打印浮点型数据时，会去掉多余的零，至多保留六位有效数字（不同于%e的默认保留小数点后6位）//2.当%g用于打印超过6位的浮点型数据时，因为精度问题，%f不得不输出一个不精确的超过六位的数字，//  %e也是同样，而%g此时会选择%e格式进行输出，并且按第一条要求，去掉多余的零，并且四舍五入到6位数字。//3.当一个数字的绝对值很小的时候，要表示这个数字所需要的字符数目就会多到让人难以接受。//  例如，把π*10^-10写作0.00000000000314159就会显得非常丑陋不雅，反之，如果我们写作3.14159e-10，就不但简洁而且易读好懂。//  当指数是-4时，这两种表现形式大小相同。对于比较小的数值，除非该数的指数小于或者等于-5，%g才会采用科学计数（%e的格式）进行输出。

/* 找出一个int数组中的最大数 */#include <stdio.h>int max (int[], int);int main (void){    int array[] = {4, 3, 6, 2, 8, 6};    printf("%d\n",sizeof(array)/sizeof(int));    printf ("The max is: %d\n", max (array, sizeof(array)/sizeof(int)));    return 0;}int max (int a[], int n){    int i, max;
　　//for (i = 0, max = a[0]; i < (sizeof(a)/sizeof(int)); i++)//sizeof(a)不能反应sizeof(array)的大小，所以处理数组的函数应该包括一个数组大小的参数    for (i=0,max=a[0]; i<n; i++)    //n反应出sizeof(array)的大小        max = (max>a[i]) ? max : a[i];         return max;}

#include <stdio.h>int MaxIndex (double * ptr, int n);int main (void){    double array[] = {4.3, 5.3, 2.6, 9.2, 2.8, 3.6};    printf ("The max number‘s index is: %d\n", MaxIndex(array, sizeof(array)/sizeof(double)));    return 0;}int MaxIndex (double a[], int n){    int i;    double max;    for (i=0,max=a[0]; i<n; i++)        max = (max>a[i]) ? max : a[i];    for (i=0; max!=a[i]; i++)        continue;    return i;}

/*求数组中最大值最小值间的差值*/#include <stdio.h>double gap (double *, int);int main (void){    double array[] = {4.3, 5.3, 2.6, 9.2, 2.8, 3.6};    printf ("The gap between max and min is: %g\n", gap (array, sizeof(array)/sizeof(double)));        return 0;}double gap (double a[], int n){    int i;    double max, min;    for (i = 1, max = a[0], min = a[0]; i < n; i++)    {        if (max < a[i])  max = a[i];        if (min > a[i])  min = a[i];    }    return (max - min);}

/*复制二维数组*/#include <stdio.h>#define ROWS 2#define COLS 3void copy_2d (double source[][COLS], double target[][COLS], int);    //复制二维数组，注意声明处理二维数组的函数时要给出子数组大小信息void copy_1d (double a1[], double a2[], int n);                        //复制一维数组int main (void){    int i, j;    double source[ROWS][COLS] = {{1, 2, 3}, {4, 5, 6}};    double target[ROWS][COLS] = {0};    copy_2d (source, target, ROWS);    for (i=0; i<ROWS; i++)    {        for (j = 0; j < COLS; j++)            printf ("%g\t", target[i][j]);        printf ("\n");    }    return 0;}void copy_2d (double (*source)[COLS], double target[][COLS], int n){    int i;    for (i = 0; i < n; i++)        copy_1d ( *(source + i), target[i], COLS);    //*(source + i), target[i]都是double(* )[3]型}void copy_1d (double a1[], double *a2, int n){    int i;    for (i = 0; i < n; i++)        a2[i] = a1[i];}

/*一维数组函数调用，复制数组中某部分的元素们*/#include <stdio.h>void copy (double *, double *, int);int main (void){    double source[] = {1.1, 2.2, 3.3, 4.4, 5.5, 6.6, 7.7};    double target[3] = {0};    printf ("Before operation:\n");    printf ("source: \t%g\t%g\t%g\t%g\t%g\t%g\t%g\n",         source[0], source[1], source[2], source[3], source[4], source[5], source[6]);    printf ("target: \t%g\t%g\t%g\n",         target[0], target[1], target[2]);        copy (source + 2, target, 3);    //函数在这里哦，怕你看得心烦不想找    printf ("\nAfter operation:\n");    printf ("source: \t%g\t%g\t%g\t%g\t%g\t%g\t%g\n",         source[0], source[1], source[2], source[3], source[4], source[5], source[6]);    printf ("target: \t%g\t%g\t%g\n",         target[0], target[1], target[2]);        return 0;}void copy (double *p1, double *p2, int n){    int i;    for (i = 0; i < n; i++)        *(p2 + i) = *(p1 +i);}

#include <stdio.h>#define COLS 5void copy (double (*)[COLS], double (*)[COLS], int);void display (double (*)[COLS], int);int main (void){    double source[3][COLS] =     {        {1.1, 2.2, 3.3, 4.4, 5.5},        {6.6, 7.7, 8.8, 9.9, 10.10},        {11.11, 12.12, 13.13, 14.14, 15.15}    };    double target[6][COLS] = {0};    copy (source, target, 3);    puts ("source:");    display (source, 3);    puts ("\ntarget:");    display (target, 5);    return 0;}void copy (double (*source)[COLS], double target[][COLS], int rows)    //double (*source)[COLS]等同于double source[][COLS]{    int i, j;    for (i = 0; i < rows; i++)    {        for (j = 0; j < COLS; j++)            target[i][j] = source[i][j];    }}void display (double (*p)[COLS], int rows){    int i,j;    for (i = 0; i < rows; i++)    {        for (j = 0; j < COLS; j++)            printf ("%g\t", p[i][j]);        printf ("\n");    }        }

/*两数组对应元素相加放在第三个数组中*/#include <stdio.h>void add (double [], double *, double [], int n);int main (void){    /**************声明和初始化两个源数组和一个目标数组*****************/    double source1[5] = {1.1, 2.2, 3.3, 4.4, 5.5};    double source2[5] = {6.6, 7.7, 8.8, 9.9, 10.10};    double target[5] = {0};        /**************执行两个源数组相加并且赋值给目标数组*****************/    add (source1, source2, target, sizeof(source1)/sizeof(double));    /**************显示两个源数组和目标数组*****************/    printf ("source1: \t%g\t%g\t%g\t%g\t%g\n",source1[0],source1[1],source1[2],source1[3],source1[4]);    printf ("source2: \t%g\t%g\t%g\t%g\t%g\n",source2[0],source2[1],source2[2],source2[3],source2[4]);    printf ("target:  \t%g\t%g\t%g\t%g\t%g\n",target[0],target[1],target[2],target[3],target[4]);    return 0;}/**************定义函数：两个源数组相加并且赋值给目标数组*****************/void add (double source1[], double source2[], double target[], int n){    int i;    for (i = 0; i < n; i++)        target[i] = source1[i] + source2[i];}

/*二维数组元素翻一番*/#include <stdio.h>#define ROWS 3#define COLS 5void twice (double (*)[COLS], int);void display (double (*)[COLS], int);int main (void){    /*******************声明和初始化二维源数组********************/    double array[ROWS][COLS] =     {        {1.1, 2.2, 3.3, 4.4, 5.5},        {6.6, 7.7, 8.8, 9.9, 10.10},        {11.11, 12.12, 13.13, 14.14, 15.15}    };    /*******************显示源数组，对源数组执行翻倍，显示翻倍后的数组********************/    puts ("before: \n");    display (array, 3);    twice (array, 3);    puts ("\nafter:\n");    display (array, 3);    return 0;}/*******************定义函数：翻倍源数组********************/void twice (double (*p)[COLS], int rows){    int i, j;    for (i = 0; i < rows; i++)    {        for (j = 0; j < COLS; j++)            *(*(p+i)+j) = 2 *(*(*(p+i)+j));    }}/*******************定义函数：显示二维数组********************/void display (double (*p)[COLS], int rows){    int i, j;    for (i = 0; i < rows; i++)    {        for (j = 0; j < COLS; j++)            printf ("%g\t", p[i][j]);        printf ("\n");    }}

/*rain.c 针对若干年的降水量数据，计算年降水总量、年降水平均量，以及月降水平均量*/#include <stdio.h>#define MONTHS 12#define YEARS 5void RainPerYear(const float (*)[12]);    //注意使用const定义形参（因为实参都用const定义了，更别说形参了）void RainAveMon(const float (*)[12]);int main (void){    /***********把数组初始化为2000到2004年的降水量数据****************/    const float rain[YEARS][MONTHS] = //const声明和初始化数组可保护数据    {        {4.3, 4.3, 4.3, 3.0, 2.0, 1.2, 0.2, 0.2, 0.4, 2.4, 3.5, 6.6},        {8.5, 8.2, 1.2, 1.6, 2.4, 0.0, 5.2, 0.9, 0.3, 0.9, 1.4, 7.3},        {9.1, 8.5, 6.7, 4.3, 2.1, 0.8, 0.2, 0.2, 1.1, 2.3, 6.1, 8.4},        {7.2, 9.9, 8.4, 3.3, 1.2, 0.8, 0.4, 0.0, 0.6, 1.7, 4.3, 6.2},        {7.6, 5.6, 3.8, 2.8, 3.8, 0.2, 0.0, 0.0, 0.0, 1.3, 2.6, 5.2}    };    /************计算和显示年总降水量*****************/    RainPerYear(rain);    /***********计算和显示月平均降水量****************/    RainAveMon(rain);    return 0;}/************定义函数：计算和显示年总降水量*****************/void RainPerYear(const float (*rain)[12]){    int years,months;    float totyear;    printf("YEAR  RAIN:\n");    for(years=0; years<YEARS; years++)    {        for(months=0,totyear=0; months<MONTHS; months++)            totyear += *(*(rain+years)+months);        printf("%4d %4.2f\n",2000+years,totyear);    }}    /***********定义函数：计算和显示月平均降水量****************/void RainAveMon(const float (*rain)[12]){    int years,months;    float totmonth;    printf("\nmonth average:\n");    printf("JUN  FEB  MAT  APR  MAY  JUN  JLY  AUG  SPT  OCT  NOV  DEC:\n");    for(months=0; months<MONTHS; months++)    {        for(years=0,totmonth=0; years<YEARS; years++)            totmonth += *(*(rain+years)+months);        printf("%-4.1f ",totmonth/5);    }    printf("\nDone!\n");}    

/*键盘输入储存3*5数组数据，计算平均，找最值，显示*/#include <stdio.h>#define COLS 5void store (double p[][COLS], int row);double average_row (double p[], int n);double average_total (double p[][COLS], int row);double max(double p[][COLS], int row);//void display (double p[][COLS], int rows, double average0, double average1, double average2, double average_total, double max);int main (void){    int i;    double array[3][COLS];    double ave_row[3];    double ave_tot;    double max_tot;    /**********************读入array的3个数集，每个数集5个数据***********************/    store (array, 3);    /****************************计算每个数集的平均值********************************/    for(i=0; i<3; i++)        ave_row[i] = average_row (array[0],5);    /****************************计算整个数集的平均值********************************/    ave_tot = average_total(array, 3);    /****************************找出整个数集的最大值********************************/    max_tot = max (array, 3);        /**********************************输出结果**************************************/    printf("ave_row[0],ave_row[1],ave_row[2] are: %g %g %g\n",ave_row[0],ave_row[1],ave_row[2]);    printf("ave_total is %g:\n", ave_tot);    printf("max is %g:\n", max_tot);    printf("Done!\n");    return 0;}/**********************定义函数：读入array的3个数集，每个数集5个数据***********************/void store (double p[][COLS], int row){    int i, j;    printf ("Please enter 15 double numbers:\n");    for (i = 0; i < row; i++)    {        for (j = 0; j < COLS; j++)            scanf ("%lf", *(p+i)+j);    // *(p+i)+j等同于&p[i][j]    }    puts ("Now you have finished.\n");}/****************************定义函数：计算每个数集的平均值********************************/double average_row (double p[], int n){    int i;    double sum;    for (i=0,sum=0; i<n; i++)        sum += p[i];    return sum/n;}/****************************定义函数：计算整个数集的平均值********************************/double average_total(double p[][COLS], int row){    int i,j;    double total=0;    for(i=0; i<row; i++)    {            for(j=0; j<COLS; j++)            total += p[i][j];    }    return total/row/COLS;}/****************************定义函数:找出整个数集的最大值********************************/double max(double p[][COLS], int row){    int i,j;    double m=p[0][0];    for(i=0; i<row; i++)        for(j=0; j<COLS; j++)            m = (m>p[i][j]) ? m : p[i][j];    return m;}