weixin_44876693的博客

#include<stdio.h>int main()//鞍点{ int arr[4][4] = { 1,7,4,1,4,8,3,6,1,6,1,2,0,7,8,9 }; int i = 0, j = 0, max = 0, col = 0,k=0, flag = 1; for (i = 0; i < 4; i++) { max = arr[i][0]; for (j = 0; j < 4; j++) { if (arr[i][j] > max).

#include<stdio.h>void func(int x){if(x==1) printf("%d",x%2);else {func(x/2);printf("%d",x%2);}x%2}int main(){int n=0;scanf("%d",&n);func(n);}

//有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月//后每个月又生一对兔子，假如兔子都不死，问每个月的兔子总数为多少？//1 1 2 5 8 13 21#include<stdio.h>int main(){ int i = 0; long int fib1=1, fib2=1; for (i = 0; i <= 10; i++) { printf("%-3d %-3d ", fib1, fib2); if ((i+1) % 2 == 0..

#include <stdio.h>int main(){ int a=0, b=0; int t; scanf("%d %d", &a, &b); int x = a, y = b; //最大公因数，辗转相除法 while (b) { t = a % b; a = b; b = t; } //最小公倍数 t = x * y / a; printf("公因数为：%d 公倍数为：%d", a,t);}...

//将一个正整数分解质因数。例如：输入90,打印出90=233*5。//对n进行分解质因数，应先找到一个最小的质数k，然后按下述步骤完成：//(1)如果这个质数恰等于n，则说明分解质因数的过程已经结束，打印出即可。//(2)如果n<>k，但n能被k整除，则应打印出k的值，并用n除以k的商, 作为新的正整数你n,//重复执行第一步。//(3)如果n不能被k整除，则用k + 1作为k的值, 重复执行第一步。#include<stdio.h>int main(){ i.

#include<stdio.h>int main(){ int a[5][5]; int i,j; int *p=a[0]; for(i=0; i<5*5; i++) p[i]=i+1; for(i=0; i<5; i++) { for(j=0; j<=i; j++) { printf("%d ",a[i][j]);} printf("\n"); }...

一个5x5的数组将每行最大值变成0将剩余数据输出#include<stdio.h>#include<time.h>#include<stdlib.h>int main(){ int a[5][5]; int b[5]; srand((unsigned)time(NULL)); int *p=a[0]; int i,j; for(i=0; i<=5*5; i++) *(p+i)=rand().

将无序数列降序将10插入数组中#include<stdio.h>int main(){ int a[9] = {14,20,5,6,7,8,11,12}; int temp1, temp2, number=10, end; int i,j,key; for(i=1; i<8; i++)//用个插入排序 { key=a[i]; j=i-1; while(j>=0&&a[.

#include<stdio.h>int main(){ int a[3][3]= {1,2,3,4,12,13,14,21,23}; int i,j,s=0,sum=0; for(i=0; i<3; i++) { for(j=0; j<3; j++) { if(i==j) s=s+a[i][j]; if(i+j==2) sum=sum

将二维数组看成一维数组进行冒泡排序int (*p)[5]，是行指针int *p是int单元的列指针#include<stdio.h>int main(){ int i=0; int j=0; int tmp; int arr[3][5]= {99,98,97,96,95,94,93,92,91,90,89,87,86,84,21}; int *p=0; p=arr;//列指针=行指针 int max=arr[0][0]; .

#include<stdio.h>#include<string.h>int huiwen(char *str){ int left=0; int right=strlen(str)-1; while(left<right) { if(str[left]!=str[right]) return 1; left++; right--; } return 0;}

#include<stdio.h>#include<string.h>int main(){char str[]="zhangqilingxingconglianyexiukillfafachuzihang";char word[26]={0};int i=0;for(i=0;i<strlen(str);i++){word[str[i]-'a']++;//这个好好琢磨是这样的，str[i]如果是b，b-a就是1，word[26]都是0，//此时word[1

#include<stdio.h>#include<string.h>char *my₋str(char* dest,char * src){ if(*src=='\0')//如果src直接为空，那就返回dest return dest; char *s1=NULL;//遍历dest字符串 char *s2=NULL;//遍历src字符串 char *sign=dest;//记录dest while(*sign)

#include<stdio.h>char *kk₋str(char* dest,char * src){ if(*src=='\0')//如果src直接为空，那就返回dest return dest; char *s1=NULL;//遍历dest字符串 char *s2=NULL;//遍历src字符串 char *sign=dest;//记录dest while(*sign) { s1=sign;//每回都让s

#include<stdio.h>void delete(char *str){ char *search=str;//遍历整个字符串 char *record=str;//记录非空格字符串 while(*search) { if(*search!=' ')//如果不是空格，就往record里加 { *record=*search; record++; }

#include<stdio.h>int main(){ int arr[]= {1,2,3,4,5,6,7}; int tmp,i; int left=0,right=sizeof(arr)/sizeof(arr[0])-1; int n=right; while(left<right) { tmp=arr[left]; arr[left]=arr[right]; arr[right]=

#include<stdio.h>char* kk₋strcat(char * dest,char * src);int kk₋strlen(char * str)//模拟实现strlen{ int count=0; while(*str++) count++; return count;}char *kk₋strcpy(char *dest,char *src)//模拟实现strcpy{ char* ret=dest; while(*des

#include<stdio.h>void search(float(*p)[4],int n);void average(float *p,int n){ int i=0; float sum=0.0,aver=0.0; for(i=0; i<n; i++) sum+=p[i]; aver=sum/n; printf("sum=%5.2f,aver=%5.2f\n",sum,aver);}void search(float

#include<stdio.h>void reading(int * pa,int m){ int i; for(i=0; i<m; i++) scanf("%d ",pa+i);//pa+i→&pa[i]}void print(int *pa,int m){ int i; for(i=0; i<m; i++) printf("%d ",pa[i]);//pa[i]→*(pa+i)}int mai

#include<stdio.h>int f(int * pa,int * pb)//形参为解引用a和b{ int tmp=0; tmp=*pa; *pa=*pb; *pb=tmp;}int main(){ int i; int a,b; scanf("%d %d",&a,&b); f(&a,&b);//实参为a和b的地址 printf("a=%d,b=%d",a,b); r

产生20个3位正整数使用插入法排序再用二分查找找到输入的数x输出其下标，找不到输出-1#include<stdio.h>#include<stdlib.h>#include<time.h>int main(){ int a[20],i,j,x; int key; srand(time(0)); for(i=0; i<20; i++)//产生三位数[m,n]，[100,999] { //m+rand()%.

#include<stdio.h>#include<stdlib.h>#include<time.h>int main(){ int a[20],i,j; int flage=1,t; srand((unsigned)time(0)); for(i=0; i<20; i++)//产生三位数[m,n]，[100,999] {//m+rand()%(n-m+1) a[i]=flage*(100+rand()

#include<stdio.h>int binary₋search(int a[],int k,int s){ int left=0; int right=s-1; int mid; while(left<=right) { mid=(left+right)/2; if(a[mid]<k) { left=mid+1; } else if(a[mid

谭浩强C程序设计page109输入一个数求位数按位逆序#include<stdio.h>int main(){ int num; scanf("%d",&num); int count=1; int n=num; while(n/10) { n/=10; count++; } printf("是%d位数\n",count); while(num) { n.

#include<stdio.h>int main(){ int a[3][4]= {18,24,23,84,56,66,45,7,5,9,38,43}; int row=0; int col=0; int max=a[0][0]; int i=0,j=0; for(i=0; i<3; i++) { for(j=0; j<4; j++) { if(a[i][j]&

#include<stdio.h>int main(){ int a[2][3]= {1,2,3,4,5,6}; int b[3][2]; int i=0; int j=0; printf("array a is：\n"); for(i=0; i<2; i++) { for(j=0; j<3; j++) { printf("%5d",a[i][j]); b[j][

题目要求:编写水仙花函数，将水仙花存入数组a再将数组a倒序存入数组b最后打印数组a和b#include<stdio.h>#include<math.h>int isShui( int number );int main(){ int m=0,x=0,i,a[100],b[100]; for(i=100; i<1000; i++) { if ( isShui(i) ) { a[x++]=i; .

#include<stdio.h>int kk₋pow(int n,int k);int isShui( int number );int main(){ int m=0,x=0,i,a[1000]; for(i=100; i<1000; i++) { if ( isShui(i) ) { //为水仙花时返回1时将此时i的值存入数组 a[x++]=i; } } for(i=0; i&l

#include<stdio.h>#include<string.h>int main(){ char a[100]; int i; gets(a); for(i=0; i<strlen(a); i++) { if(a[i]>='a'&&a[i]<='z') a[i]-=32; else if(a[i]>='A'&&a[

#include<stdio.h>/*米是(foot+inch/12)*0.3048。用户输入的是厘米数别忘了1英尺等于12英寸。输入格式：输入在一行中给出1个正整数，单位是厘米。输出格式：输出这个厘米数对应的英尺和英寸的整数值，中间用空格分开。*/int main(){int cm=0;scanf("%d",&cm);int foot=cm/30.48;int inch=(cm/30.48-foot)*12; printf("%d %d",foot,i

可使用回调函数优化#include<stdio.h>int min(int x,int y);int mul(int x,int y);int div(int x,int y);/*计算2个正整数的和、差、积、商并输出。输入格式：输入在一行中给出2个正整数A和B。输出格式：“A 运算符 B = 结果”顺序输出和、差、积、商。*/int add(int x,int y) { return x+y;}int min(int x,int y) { return

指针实现#include<stdio.h>int kk₋strlen(char * str){//用指针实现char *tmp=str;while(*str++);//str地址++再解引用，如果读到0就跳出return str-tmp-1;}int main(){ char str[]="killfafa"; int ret=kk₋strlen(str); printf("%d",ret); return 0;}计数器实现#inc.

#include<stdio.h>char *kk₋strcpy(char* dest,const char*src) { char* ret=dest; while(*dest++=*src++); return ret;}int main(){ char arr[]="abcdef"; char *p="ABCDE"; kk₋strcpy(arr,p); printf("%s",arr); return 0;}.

#include<stdio.h>int fun(int x){if(x==1)return 1;elsereturn x*fun(x-1);}int main(){int n=0;scanf("%d",&n);if(n<=0)printf("the number is wrong");else printf("%d!：%d",n,fun(n)); return 0;}

#include<stdio.h>int main( ){ int i,j; int n,m,a[100][100]; scanf("%d %d",&n,&m); for(i=0;i<n;i++) { for(j=0;j<m;j++) scanf("%d",&a[i][j]); } for (int i = 0; i < 3; ++i) {

#include<stdio.h>double my₋pow(int n,int k) { if(k<0) //为负次方时 return (1.0/(my₋pow(n,-k))); //2的-1次方=1/2¹ else if(n==0) //0次方为1 return 1; else return n*my₋pow(n,k-1); //递归}int main(){ int a,b; scanf

#include<stdio.h>int max2₋digital(int a,int b);//穷举法int max3₋digital(int a,int b);//递归int min₋digital(int a,int b);//最小公倍数int max1₋digital(int a,int b)//辗转相除法求最大公因数{ int c; while(b) { c=a%b; a=b;

#include<stdio.h>int my₋strlen(char *plen);char reverse(char *pstr,char *pend) { char tmp; while(pstr<=pend) { tmp=*pstr; *pstr=*pend; *pend=tmp; pstr++; pend--; }

#include<stdio.h>int main(){ int arr[10][10]= {0}; int i=0,j=0; for(i=0; i<10; i++) { for(j=0; j<=i; j++) { if(j==0||i==j) //输出外层1

#include<stdio.h>int main(){int i=0;int j=0;int n=0;scanf("%d",&n);for(i=1;i<=n;i++){for(j=1;j<=n+i-1;j++){//n+i-1：每行*的位置if(j==n-i+1||j==n+i-1)//n+i-1：每行最后*的位置printf("*");elseprintf(" ");}printf("\n");}//————————↑打印上半部分△ for(i=1;i<n;i