C语言程序输入有一个英文句子,将每个单词的第一个字母改为大写.

#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i

给,最简代码：#includevoidmain(){inta,b,c,max;scanf("%d%d%d",&a,&b,&c);max=(a>b?a:b)>c?(a>b?a:b):c;printf("

intmain(){intcountA_Z[26];intcounta_z[26];charc;inti;for(i=0;i='a'&&c='A'&&c

printf("请输入一个整数%d:",x)；printf("最大值为：%d\n",x)；printf("最小值为：%d",z)；我不明白,根据你的题意,应该用多分支的if语句的格式,即：if(表达式

#include"stdio.h"voidmain(){inta,b;inti,n;intflag=1;printf("请输入上限：");scanf("%d",&a);printf("请输入下限：")

#include#includeintmain(){charstr[100];inti;gets(str);for(i=0;i

if(k==0&&a[i]122)这句有逻辑错误.a[i]不可能同时小于97和大雨122.后面那个&&应该是||.应该是这句引起的.

给一个最粗糙的版本,能实现#includemain(){\x09charstr[50];\x09inti,len;\x09printf("inputasentence:\n");\x09gets(st

main(){inta=0,i;scanf("%d",&a);for(i=0;i

voidmain(){floatr,s,l;floatpi=3.1415;printf("输入一个圆的半径:\n");scanf("%f",&r);if(

#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}

#includemain(){charstring[81];inti,num=0,word=0;charc;gets(string);for(i=0;(c=string[i])!='\0';i++)/

#include"stdio.h"intmain(){inta,b,c,tmp,max;printf("inputa,b,c:");scanf("%d%d%d",&a,&b,&c);if(a>b&&a

可以用sprintf(str,"%d",num);将数字输出为字符串,再用strlen(str)检查其长度：intnulen(intnum){charstr[10];sprintf(str,"%d",

#includevoidmain(){inta;printf("请输入一个三位数：");scanf("%d",&a);printf("各位数上的乘积为：%d\n",(a/100)*(a%10)*(a/

#include"stdio.h"main(){char*s;intcnt=0;printf("pleaseenterthesentence\n");gets(s);while(*s){if(*s==

感觉像是ACM的题.N

#include <stdio.h>char* dg(char* instr, char* outstr, char* 

QuickBasic

句子【Iamyourfather】有单词4个.Pressanykeytocontinue#include"stdio.h"#include"string.h"intmain(){inti,j,k=0