编程珠玑习题：数字重复出现位图排序

内存限制，数字重复出现位图排序，每个数字限定最多出现10次

#include<iostream>#include<fstream>#include<string>#include"time.h"using namespace std;#define N 10000#define NGROUP 5//一个数需要保存两个信息，1bit数值信息，4bit个数信息#define BITSTEP 32#define MASK 0x1F#define shift 5#define shiftnum 3//4bit保存一个数字出现的次数#define MEMORY 100//内存限制#define N_S (MEMORY*8/NGROUP) //按照数字范围处理，每趟能处理的数字范围，即每趟能处理多少个数#define COUNT (N/N_S+1) //需要多少趟//#define COUNT(N,MEMORY) ((N)/((MEMORY) * 8) + 1)//#define NNUM_S (NNUM/COUNT)int bit[N_S/BITSTEP+1];//保存数字int bitnum[N_S*4/BITSTEP+1];//保存数字出现的次数/****************数字没出现一次计数加1，当出现次数超过15次时，报错*/void add(int i){    int tmp=bitnum[i>>shiftnum]&(0xF<<((i&0x7)*4));tmp=tmp>>(((i&0x7)*4));if(tmp==0xF){cerr<<"num over flow:  "<<i<<endl;exit(1);}bitnum[i>>shiftnum]+=(1<<((i&0x7)*4));cout<<"num   "<<i<<"count  "<<bitnum[i>>shiftnum]<<endl;}/************************返回对应数字i出现的次数*/int num(int i){ int tmp=bitnum[i>>shiftnum]&(0xF<<((i&0x7)*4)); return tmp>>((i&0x7)*4);}void clr(int i,int * bit_temp){bit_temp[i>>shift]&=~(1<<(i&MASK));}void set(int i){bit[i>>shift]|=(1<<(i&MASK));}int test(int i){int tmp=bit[i>>shift]&(1<<(i&MASK));return tmp>>(i&MASK);}int main(void){    string ifile,ofile;ifstream infile;ofstream outfile;int a;cin>>ifile;cin>>ofile;//cout<<COUNT(N,MEMORY)<<endl;//cout<<(N/2)<<endl;//NN=N/COUNT;    cout<<COUNT<<endl;    time_t str_time=time(&str_time);char * str=ctime(&str_time);//cout<<"str_time"<<str<<endl;infile.open(ifile.c_str());    outfile.open(ofile.c_str());if(!infile){cerr<<"error: can not open file :"<<ifile<<endl;return -1;}if(!outfile){cerr<<"can not open outfile :"<<ofile<<endl;return -1;}  for(int k=0;k<COUNT;k++){cout<<"round "<<k<<endl;infile.clear();//清掉EOFinfile.seekg(0,ios::beg);//回到文件头outfile.seekp(0,ios::cur);for(int i=0;i<N_S;i++){ //位图向量与位图计数向量均清零//infile<<i<<endl;   clr(i,bit); }for(i=0;i<N_S*4;i++){clr(i,bitnum);}    while(infile>>a){int count_num=0;//cout<<a<<endl;if(a>=k*N_S&&a<N_S+k*N_S){cout<<"num  "<<a<<endl;//cout<<a<<endl;a=a-N_S*k;//cout<<a<<endl;    set(a);add(a);}//cout<<bit[a>>shift]<<endl;}time_t net_time=time(&net_time);char * net=ctime(&net_time); double difnet=difftime(str_time,net_time); //cout<<"round"<<k<<endl; //cout<<"net_time"<<net<<difnet<<endl;for(i=0;i<N_S;i++){int num_tmp=num(i);cout<<"test!!!num   "<<i<<"num_count  "<<num_tmp<<endl;for(int j=0;j<num_tmp;j++){   if(test(i)==1)//cout<<test(i)<<endl;    outfile<<i+k*N_S<<endl;}}}     time_t end_time=time(&end_time);  char * end=ctime(&end_time); double dif=difftime(str_time,end_time);//double end_time=_getsystime(&t);    // cout<<"end_time"<<end<<dif<<endl;}