给40亿个不重复的unsigned int的整数，没排过序的，然后再给一个数，如何快速判断这个数是否在那40亿个数当中？

总共有4*10^9个数字，如果直接放在内存中，需要的内存量是

4*10^9*4bytes(一个unsigned int为4bytes) ~= 16GBytes内存

用bitmap表示就是4*10^9bits~=512MBytes=2^32 = 4 294 967 296 ，所以索性就分配512M内存，然后依次将这个40亿不重复的数字添加到bitmap中，默认为0，添加后为1。最后直接判断要查找的数字的bitmap值，如果为0，则表示没有，如果为1，则表示有。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define TOTAL_DATA_SIZE 300 //4000000000  /*40亿个不重复的unsigned int整数大数据*/
//#define MAX_BUFFER_SIZE 536870912   /*512MB*/
typedef char bitmap_type;
#define BITMAP_BITS (sizeof(bitmap_type)*8)
#define MAX_BUFFER_SIZE ((TOTAL_DATA_SIZE+BITMAP_BITS-1)/BITMAP_BITS)

#define bitmap_set(b,v) (b[(v)/BITMAP_BITS] |= (1 << ((v)%BITMAP_BITS)))
#define bitmap_clear(b,v) (b[(v)/BITMAP_BITS] &= ~(1 << ((v)%BITMAP_BITS)))
#define bitmap_isset(b,v) (b[(v)/BITMAP_BITS] & (1 << ((v)%BITMAP_BITS)))
#define bitmap_zero(b) memset(b,0,sizeof(bitmap_type)*MAX_BUFFER_SIZE)

#define DATA_FILE_NAME "data.txt"


int main(int argc, char*argv[])
{
    bitmap_type *bitmap = NULL;
    FILE *fp;
    unsigned int num;
    bitmap = (bitmap_type*)malloc(sizeof(bitmap_type)*MAX_BUFFER_SIZE);
    if(bitmap == NULL) {
        printf("bitmap malloc error\n");
        return -1;
    }
    bitmap_zero(bitmap);
    //memset(bitmap,0,sizeof(bitmap_type)*MAX_BUFFER_SIZE);
    fp = fopen(DATA_FILE_NAME,"rb");
    if(fp == NULL) {
        printf("fopen file %s error\n",DATA_FILE_NAME);
        return -1;
    }
    while(fscanf(fp,"%d",&num) != EOF) {
        bitmap_set(bitmap,num);
    }
    if(bitmap_isset(bitmap,2)) {
        printf("2 is set in the bitmap\n");
    }
    if(bitmap_isset(bitmap,222)) {
        printf("222 is set in the bitmap\n");
    }else {
        printf("222 is not in the bitmap\n");
    }
    if(bitmap_isset(bitmap,333)) {
        printf("333 is set in the bitmap\n");
    }else {
        printf("333 is not in the bitmap\n");
    }


    return 0;
}

[root@localhost c_language]# more data.txt

1

2

3

222

5

6

7

8