用C语言设计一个函数，取出字符串“A/BB/CCC"中的“A”“B”“C”
\u5728\u7ebf\u7b49 C\u8bed\u8a00\u4e2d\uff0c\u6c42\u5b57\u7b26\u4e32\u5b50\u4e32\u7684\u51fd\u6570\u5728\u7eafC\u4e2d\u597d\u50cf\u6ca1\u6709\uff0c\u4f46\u662fC++\u4e2d\u5f88\u5bb9\u6613\uff0c\u4f60\u53ef\u4ee5\u53bb\u67e5\u4e00\u4e0b\uff0c\u7f51\u4e0a\u6709\u5f88\u591a\u7eafC\u5b9e\u73b0\u7684\u5b50\u4e32\u6c42\u6cd5
string\u7684substr\u51fd\u6570\u662fC++\u7684\uff0c\u662fSTL\u5e93\u4e2d\u7684\uff0c\u800c\u975e\u7eafC\u7684\u5b9e\u73b0\uff0c\u7eafC\u7684\u6807\u51c6\u5e93\u4e2d\u6ca1\u6709\u76f4\u63a5\u83b7\u5f97\u5b50\u4e32\u7684\u65b9\u6cd5\uff0c\u81ea\u5df1\u5199\u5f88\u5bb9\u6613
\u53ef\u4ee5\u7684\u3002c++\u662f\u5411\u4e0b\u5bf9C\u8bed\u8a00\u517c\u5bb9\u7684\u3002C\u8bed\u8a00\u4e2d\u6240\u6709\u7684\u7279\u6027\u6216\u51fd\u6570\u8fd8\u6709\u64cd\u4f5c\u7b26\u90fd\u53ef\u5728C++\u4e2d\u4e0d\u7ecf\u8fc7\u4fee\u6539\u76f4\u63a5\u4f7f\u7528\u3002\u4e0d\u540c\u7684\u662f\uff0cC++\u7684\u6807\u51c6\u51fd\u6570\u5e93\uff08STL\uff09\u53ef\u80fd\u63d0\u4f9b\u4e86\u66f4\u597d\u7684\u5b9e\u73b0\u65b9\u5f0f\u3002\u697c\u4e3b\u53ef\u4ee5\u5927\u80c6\u5728C++\u4e2d\u4f7f\u7528C\u7684\u51fd\u6570\uff0c\u4e0d\u7528\u62c5\u5fc3\u7684\u3002
int main(){
char a[]="A/BB/CCC";
for(int i=0;i<strlen(a);i++)
{
if(a[i]!='/')
printf("%c",a[i]);
}
}
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