c语言数位拆分，从键盘上输入一个4位数的整数n，编写程序将其拆分为两个2位数的整数a和b 四位数拆分c语言设计
\u7528C\u8bed\u8a00\u5982\u4f55\u5c06\u4ece\u952e\u76d8\u8f93\u5165\u56db\u4f4d\u6570,\u5206\u89e3\u62104\u4e2a\u6570\u5b57\u5047\u8bbe\u4f60\u8f93\u5165\u7684\u6570\u4e3ax;
unsigned char s[4];
s[0]=x/1000;//\u56db\u4f4d\u6570\u7684 \u5343\u4f4d
s[1]=x/100%10;\u56db\u4f4d\u6570\u7684 \u767e\u4f4d
s[2]=x/10%10;\u56db\u4f4d\u6570\u7684 \u5341\u4f4d
s[3]=x%10;\u56db\u4f4d\u6570\u7684 \u4e2a\u4f4d
\u8f93\u51fa\u65b0\u76842\u4e2a\u56db\u4f4d\u6570\u7528\uff08a\u548cb\uff09\u8868\u793a\uff01 \u628a\u4ed6\u4eec \u5343\u767e\u5341\u4e2a \u7684\u4f4d\u7f6e\u6362\u4e0b\u884c\u4e86\uff0c\u5047\u5982\u8f93\u5165\u7684x=1234,\u90a3\u4e48\uff1a
a=s[1]*1000+s[0]*100+s[2]*10+s[3];//a=2134\u662f\u4e00\u4e2a\u65b0\u7684\u56db\u4f4d\u6570
b=ss[2]*1000+s[1]*100+s[3]*10+s[0];//a=3241\u662f\u4e00\u4e2a\u65b0\u7684\u56db\u4f4d\u6570
\u6700\u540eprint\u51fa\u6765\u5c31\u884c\u4e86\u3002
\u56e0\u4e3aC\u8bed\u8a00\u597d\u4e45\u4e0d\u7528\u8f93\u5165\u8f93\u51fa\u7684\u8bed\u6cd5\uff0c\u5fd8\u8bb0\u4e86\uff0c\u672c\u60f3\u5e2e\u4f60\u5199\u4e2a\u5b8c\u6574\u7684\u7a0b\u5e8f\u7684\u3002\u4f46\u53ea\u80fd\u5e2e\u4f60\u5230\u8fd9\u4e86\uff01\u5168\u624b\u6253\uff01\u5e0c\u671b\u5e2e\u5230\u4f60\u4e86\uff01
#include
void main()
{
int a,b,c,d,e,t;
printf("Please input a number:\n");
scanf("%d",&t);
if(t>999&&t<10000)
{
a=t/1000;
b=t%1000/100;
c=t%100/10;
d=t%10;
e=a+b+c+d;
if(e%2==0)
printf("YES\n");
else
printf("NO\n");
}
else
printf("Invalid data\n");}
不要这些switch和case，就可以得到结果。
绛旓細op娌℃湁鍒濆鍖栵紝瀹冧笉鏄+-*/涓殑浠讳綍涓涓紝鎵浠ユ棤鏄剧ず銆備笉瑕佽繖浜泂witch鍜宑ase锛屽氨鍙互寰楀埌缁撴灉銆
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