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处理二维数组的函数#include<stdio.h>#define ROWS 3#define COLS 4void sum_rows(int ar[][COLS], int rows);void sum_cols(int [][COLS], int); // 省略形参名，没问题int sum2d(int ar[][COLS], int rows); // 另一种语法int main(){ int junk[ROWS][COLS] = { {2, 4, 6, 8},

指针与多维数组的理解#include<stdio.h>int main(){ int zippo[4][2] = { {2, 4}, {6, 8}, {1, 3}, {5, 7} }; printf(" zippo = %p, zippo + 1 = %p\n", zippo, zippo + 1); // +1意味着加两个int printf(" zippo[0] = %p, zippo[0]

处理数组的函数显示数组内的所有元素使数组内的每个元素都乘以2.5// arf.c -- 处理数组的函数#include<stdio.h>#define SIZE 5void show_array(const double ar[], int n);void mult_array(double ar[], int n, double mult);int main(vodi){ double dip[SIZE] = { 20.0, 17.66, 8.2, 15.3, 22.

指针的基本操作#include<stdio.h>int main(){ int urn[5] = { 100,200,300,400,500 }; int* ptr1, * ptr2, * ptr3; ptr1 = urn; // 把一个地址赋给指针 ptr2 = &urn[2]; printf("pointer value, dereferenced pointer, pointer address:\n"); printf("ptr1 = %p, *ptr

指针运算中的优先级#include<stdio.h>int data[2] = { 100,200 };int moredata[2] = { 300,400 };int main(){ int* p1, * p2, * p3; p1 = p2 = data; p3 = moredata; printf(" *p1 = %d, *p2 = %d, *p3 = %d\n", *p1, *p2, *p3); printf("运算之前：p1 = %p, p2 = %p,

数组元素之和#include<stdio.h>#define SIZE 10int sump(int* start, int* end);int main(){ int marbles[SIZE] = {20, 10, 5, 39, 4, 16, 19, 26, 31, 20}; long answer; answer = sump(marbles, marbles+SIZE); printf("The total number of marbles is %ld.\n",

数组元素之和#include<stdio.h>#define SIZE 10int sum(int ar[], int n);int main(){ int marbles[SIZE] = { 20, 10, 5, 39, 4, 16, 19, 26, 31, 20 }; long answer; answer = sum(marbles, SIZE); printf("The total number of marbles is %ld.\n", answer); pr

多维数组#include<stdio.h>#define MONTHS 12#define YEARS 5int main(){ const float rain[YEARS][MONTHS] = { {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

读取数字给出相应的英文月份#include<stdio.h>int main(){ int num; char *month[]={"Jan","Feb","Mar","Apr","May","Jun","Jul","Aus","Sep","Oct","Nov","Dec"}; scanf("%d",&num); printf("%s\n",month[num-1]);}

利用指针求数组的最大最小值#include<stdio.h>void minmax(int a[], int len, int *max, int *min);int main(void){ int a[] = {1,2,3,4,5,6,7,8,12,13,14,16,17,21,23,55}; int min, max; minmax(a, sizeof(a)/sizeof(a[0]), &min, &max); printf("min

指针交换变量#include<stdio.h>void swap(int *pa, int *pb);int main(){ int a = 5; int b = 6; swap(&a, &b); printf("a=%d, b=%d", a, b); return 0;}void swap(int *pa, int *pb){ int t = *pa; *pa = *pb; *pb =

构造素数表#include<stdio.h>int main(){ const int maxNumber = 25; int isPrime[maxNumber]; int i; int x; for ( i = 0; i < maxNumber; i++) { isPrime[i] = 1; } // for test printf("\t"); for ( i = 2; i < maxNumber; i++) { printf("%d\t",

数组找数字#include<stdio.h>int search(int key, int a[], int length); // 声明int main(void){ int a[] = { 2,4,6,7,1,3,5,9,11,13,23,14,32 }; int x; int loc; printf("请输入一个数字："); scanf_s("%d", &x); loc = search(x, a, sizeof(a) / sizeof(a[0])); if

数组求均值并输出大于均值的数#include<stdio.h>int main(){ int x; double sum = 0; int cnt = 0; int number[100]; // 定义一个100个元素的数组 scanf_s("%d", &x); while (x!=-1) { number[cnt] = x; sum += x; cnt++; scanf_s("%d", &x); } if (cnt>0) {

数据类型整数char、short、int、long、long longchar：1字节（8比特）short：2字节int：取决于编译器（CPU），通常的意义是“一个字”long：取决于编译器（CPU），通常的意义是“一个字”long long：8字节浮点数float、double、long double逻辑bool指针自定义类型数的范围：char<short<int<float<doublechar<short<int<float

求a的连续和输入两个整数a和n，a的范围是[0,9]，n的范围是[1,8]，求数列之和S=a+aa+aaa+...+aaa...aS=a+aa+aaa+...+aaa...aS=a+aa+aaa+...+aaa...a（n个a）。如a为2、n为8时输出的是2+22+222+...+222222222+22+222+...+222222222+22+222+...+22222222的和。输入格式：输入在一行中给出两个整数，先后表示a和n输出格式：在一行中输出要求的数列之和。输入样例：2 4输出样例：

念数字#include<stdio.h>int main(){ int x; scanf_s("%d", &x); if (x<0) { printf("fu "); x = -x; } int mask = 1; int t = x; while (t>9) { t /= 10; mask *= 10; } do { int d = x / mask; switch (d) { case 0:printf("li

约分最简分式#include<stdio.h>int main(){ int dividend, divisor; scanf_s("%d/%d", &dividend, &divisor); int a = dividend; int b = divisor; int t; while (b>0) { t = a % b; a = b; b = t; } // a为最大公约数 printf("%d/%d\n", dividend /

n项求和计算21+32+53+85+...\frac{2}{1}+\frac{3}{2}+\frac{5}{3}+\frac{8}{5}+...12​+23​+35​+58​+...#include<stdio.h>int main(){ int n; double dividend, divisor; double sum = 0.0; int i; double t; scanf_s("%d", &n); dividend = 2; divisor = 1;

#include<stdio.h>int main(){ int number, n; int inp; int finished = 0; int cnt = 0; printf("请依次输入你想猜的数字与猜的次数：\n"); scanf_s("%d %d", &number, &n); printf("请开始猜数字：\n"); do { scanf_s("%d", &inp); cnt++; if (inp<0) {

c语言 统计素数之和#include<stdio.h>int main(){ int m, n; printf("您想求哪两个数之间的素数之和，请依次输入它们："); scanf_s("%d %d", &m, &n); int i; int cnt = 0; int sum = 0; if (m==1) { m = 2; } for ( i = m; i <= n; i++) { int isPrime = 1; int k;

九九乘法表#include<stdio.h>int main(){ int n = 9; //scanf_s("%d", &n); int i, j; i = 1; while (i<=n) { j = 1; while (j<=i) { printf("%d*%d=%d", j, i, i * j); if (i*j<10) { printf(" "); } else { print

水仙花数据水仙花数是指一个NNN位数正整数（N>=3N>=3N>=3），它的每位上的数字的NNN次幂之和等于它本身。例如153=13+53+33153=1^3+5^3+3^3153=13+53+33。#include <stdio.h>int main(){ int n; printf("请输入一个数：\n"); scanf_s("%d", &n); int first = 1; int i = 1; while (i < n) { fir