用C语言编一个程序，输出1—200之间不能被2和3和4整除的数，要求一行输出10个数。 急救，c语言编程题。 编写程序,将1—100间不能被3整除的...
\u7528c\u7a0b\u5e8f\u7f16\u5199(\u8f93\u51fa100\u5230200\u4e4b\u95f4\u80fd\u88ab3\u6574\u9664\u4e0d\u80fd\u88ab4\u6574\u9664\u7684\u6570,\u5e76\u6c42\u548c)102\u80fd\u88ab3\u6574\u9664\uff0c\u6240\u4ee5\u4ece102\u5f00\u59cb\uff0c\u6b65\u957f3\u589e\u91cf\u7684\u503c\u80af\u5b9a\u90fd\u80fd\u88ab3\u6574\u9664\uff1b\u518d\u5bf9\u8fd9\u4e9b\u6570\u8fdb\u884c\u80fd\u5426\u88ab4\u6574\u9664\u7684\u8003\u5bdf\uff0c\u8f93\u51fa\u4e0d\u80fd\u88ab4\u6574\u9664\u7684\u6570\uff0c\u540c\u65f6\u7d2f\u52a0\u6c42\u548c\u3002\u4ee3\u7801\u5982\u4e0b\uff1a
#include "stdio.h"int main(int argc,char *argv[]){int s,n,t;for(t=s=0,n=102;n<200;n+=3)//\u4ece102\u5f00\u59cb\u6b65\u957f3\u589e\u91cf\u4fdd\u8bc1n\u80fd\u88ab3\u6574\u9664if(n%4){//\u8003\u5bdf\u662f\u5426\u80fd\u88ab4\u6574\u9664printf(++t%10 ? "%4d" : "%4d\n",n);//\u8f93\u51fa\u4e0d\u80fd\u88ab4\u6574\u9664\u7684\u6570s+=n;//\u5411s\u7d2f\u52a0\u6c42\u548c}if(t%10)printf("\n");printf("The sum of them is %d\n",s);//\u8f93\u51fa\u548creturn 0;}\u8fd0\u884c\u7ed3\u679c\u5982\u4e0b\uff1a
void sort(int *p1,int *p2)
{
int temp;
if(*p1 > *p2)
{
temp = *p1;
*p1 = *p2;
*p2 = temp;
}
}
void chars_num(char * str,int * upcase,int *lowcase,int * number,int *others)
{
char * p = str;
*upcase = *lowcase = *number = *others = 0;
if(str == 0)
return;
while (*p != '\0')
{
if('A' <= *p && *p <= 'Z')
(*upcase)++;
else if('a' <= *p && *p <= 'z')
(*lowcase)++;
else if('0'<= *p && *p <= '9')
(*number)++;
else
(*others)++;
p++;
}
}
int strcmp (char * p1, char * p2)
{
while ((*p1 != '\0') && (*p2 !='\0') && (*p1 == *p2) )
{
p1++;
p2++;
}
if((*p1 != '\0') && (*p2 =='\0'))
return *p1;
else if ((*p1 == '\0') && (*p2 !='\0'))
return -*p2;
else if((*p1 != '\0') && (*p2 !='\0'))
return *p1 - *p2;
else
return 0;
}
void main()
{
int a,b,c;
int temp;
char str[100];
int upcase,lowcase,number,others;
char cmp1[100],cmp2[100];
int result;
printf("\n1.\u8f93\u51653\u4e2a\u6574\u6570\uff0c\u6309\u7531\u5c0f\u5230\u5927\u7684\u987a\u5e8f\u8f93\u51fa. \n\u8bf7\u8f93\u5165\u4e09\u4e2a\u6574\u6570\uff1a");
scanf("%d %d %d",&a,&b,&c);
sort(&a,&c);
sort(&b,&c);
sort(&a,&b);
printf("\u6392\u5e8f\u4ee5\u540e\u7684\u503c\uff1a%d,%d,%d\n",a,b,c);
printf("\n2.\u7f16\u7a0b\u7edf\u8ba1\u4e00\u4e2a\u5b57\u7b26\u4e32\u4e2d\u5927\u5199\u5b57\u6bcd\u3001\u5c0f\u5199\u5b57\u6bcd\u3001\u6570\u5b57\u548c\u5176\u4ed6\u5b57\u7b26\u7684\u4e2a\u6570\u3002 \n\u8bf7\u8f93\u5165\u4e00\u4e2a\u5b57\u7b26\u4e32\uff1a");
scanf("%s",str);
chars_num(str,&upcase,&lowcase,&number,&others);
printf("\u4e0a\u9762\u5b57\u7b26\u4e32\u4e2d\u5927\u5199\u5b57\u6bcd\u4e2a\u6570\u4e3a\uff1a%d,\u5c0f\u5199\u5b57\u6bcd\u4e2a\u6570\u4e3a\uff1a%d,\u6570\u5b57\u4e2a\u6570\u4e3a\uff1a%d,\u5176\u5b83\u5b57\u7b26\u4e2a\u6570\u4e3a\uff1a%d\n",upcase,lowcase,number,others);
fflush(stdin);
printf("\n\u7528\u4e00\u4e2a\u51fd\u6570\u5b9e\u73b0\u4e24\u4e2a\u5b57\u7b26\u4e32\u7684\u6bd4\u8f83\uff0c\u5373\u81ea\u5df1\u5199\u4e00\u4e2astrcmp\u51fd\u6570\uff0c\u51fd\u6570\u539f\u578b\u4e3a\uff1aint strcmp (char * p1, char * p2)\uff1b\n");
printf("\u8bf7\u8f93\u5165\u8981\u6bd4\u8f83\u7684\u5b57\u7b26\u4e32\uff1a");
scanf("%s",cmp1);
printf("\u8bf7\u8f93\u5165\u8981\u6bd4\u8f83\u7684\u53e6\u4e00\u4e2a\u5b57\u7b26\u4e32\uff1a");
scanf("%s",cmp2);
result = strcmp(cmp1,cmp2);
printf("%s - %s = %d\n",cmp1,cmp2,result);
}
\u5206\u592a\u5c11\u4e86\uff5e\u52a0\u5206\u5427
int i;
for (i=1;i<=100;i++) if (i%2 && i%3 && i%4) printf("%d\t",i);
}
以上程序实现每行10个的办法是通过\t，无需其它任何代码，输出的结果自然每行10个，并且左边对齐。
#include <stdio.h>
void main()
{
int i,j;
j=0;
for(i=1;i<=200;i++)
{
if((i%2)&&(i%3)&&(i%4))
{
if(j==10)
{
printf("\n");
j=0;
}
printf(" %d ",i);
j++;
}
}
getch();
}
#include <stdio.h>
void main()
{
int i,sum;
for(i=1;i<=200;i++)
if(i%2!=0&&i%3!=0&&i%4=0)
{
printf("%d\t",i);
sum++;
if(sum>=10&&sum%10==0)
printf("\n");
}
}
我是初学者，估计这个好理解
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绛旓細闄2浠ュ锛岀礌鏁扮殕涓哄鏁帮紝鎵浠ュ2鍗曠嫭澶勭悊锛屼粠3寮濮嬶紝灏100浠ュ唴鐨勫鏁伴愪竴杩涜绱犳暟鑰冨療锛屾槸鍒杈撳嚭锛涓嶆槸鍒欒烦杩囥傚彟璁句竴璁℃暟鍙橀噺鎺у埗鎹㈣锛屾瘡杈撳嚭5涓礌鏁版坊鍔涓涓鍥炶溅銆備唬鐮佸涓嬶細include "stdio.h"int main(int argc,char *argv[]){ int i,n,t;pr include include main(){ int a,i,flag,coun...
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