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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:; } Followed by: Post your reply here: |
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