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HELP,都能通过就是WRONGANSWER,大家帮个忙吧,多谢#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<iostream.h>
//如何实现变换
void change(char[]);char *jk=NULL ;
char inh[100]; //char*temp[19]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int bji=1,x=1;//函数调用次数
char *temp[19];
int go=0;
void main()
{
cin>>inh;
while(!(inh[0]=='-'))
{ bji=1;
change(inh);
go=0;
for(int i=0;i<100;i++)
inh[i]=0;
// for(i=0;i<19;i++)
// temp[i]=0;
for(int g=0;g<19;g++)
temp[g]= new char[15];
cin>>inh;
x=1;
//cout<<inh;
}
}
void change( char h[])//change(tok);
{/*jk=tok;
temp[0]=inh;
temp[x++]=jk;*/
temp[0]=inh;
int token[9]={0};
for(int i=0;h[i]!=0;i++)
{
switch(h[i])
{
case '1':token[0]++;break;
case '2':token[1]++;break;
case '3':token[2]++;break;
case '4':token[3]++;break;
case '5':token[4]++;break;
case '6':token[5]++;break;
case '7':token[6]++;break;
case '8':token[7]++;break;
case '9':token[8]++;break;
}
}
//转换后输出,仅是个数在10以内的情况
char tok[100]={NULL};int p=0;//int token[];
for(i=0;i<9;i++)
{//yi wei shu qing kuang
int pivot=token[i]/10;
if(token[i]!=0&&token[i]/10==0){
tok[p++]=token[i]+'0';
tok[p++]=i+1+'0';//i+1 bu hui chao xian
}
if(pivot>=1&&pivot<10){//两位数
tok[p++]=pivot+'0';
tok[p++]=token[i]%10+'0';
tok[p++]=i+1+'0';
}
}
temp[x++]=tok;
// cout<<tok<<endl;
if(!strcmp(tok,inh)&&bji==1)cout<<inh<<" is self-inventorying "
<<endl;
else if(!strcmp(tok,h)&&bji>1)cout<<inh<<" is self-inventorying after "
<<bji-1<<" steps"<<endl;
else if(bji>1)
{
for(i=0;i<(x-2);i++){
if(!strcmp(temp[i],tok))
{//?比较对象调整
cout<<inh<<" enters an inventory loop of length "
<<bji-i
<<endl;
go=12;
break;
}
}
if(go==0)
goto P;
}
else
{
P:
bji++;
if(bji==16)
{cout<<inh
<<" can not be classified after 15 iterations"
<<endl;
goto Loop;
}
jk=tok;
//temp[0]=inh;
//temp[x++]=jk;
//strcpy(temp[0],inh);//??
//strcpy(temp[x++],jk);//x=1;CHUZHI
change(jk);
}
Loop:;
}
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