C语言实现bitmap,取两个数组的交集等操作

bitmap的基本操作：

#include <stdio.h>#include <stdlib.h>#define WORD 32#define SHIFT 5 ////移动5个位,左移则相当于乘以32,右移相当于除以32取整#define MASK 0x1F //16进制下的31#define N 10000000/* * 置位函数——用"|"操作符,i&MASK相当于mod操作 * m mod n 运算，当n = 2的X次幂的时候,m mod n = m&(n-1) */void set(int *bitmap, int i) {bitmap[i >> SHIFT] |= (1 << (i & MASK));}/* 清除位操作，用&~操作符 */void clear(int *bitmap, int i) {bitmap[i >> SHIFT] &= ~(1 << (i & MASK));}/* 测试位操作用&操作符 */int test(int *bitmap, int i) {return bitmap[i >> SHIFT] & (1 << (i & MASK));}

排序：

void sort() {int bitmap[1 + N / WORD];FILE *in = fopen("in.txt", "r");FILE *out = fopen("out.txt", "w");if (in == NULL || out == NULL) {exit(-1);}int i = 0;int m;for (i = 0; i < N; i++) {clear(bitmap, i);}while (!feof(in)) {fscanf(in, "%d", &m);printf("%d/n", m);set(bitmap, m);}printf("abnother");for (i = 0; i < N; i++) {if (test(bitmap, i)) {printf("%d/n", i);fprintf(out, "%d/n", i);}}fclose(in);fclose(out);}

求交集：

//合并两个数组，去交集void merge() {int A[] = { 1, 3, 5, 7, 9, 435, 3454, 345543, 3453455 };int B[] = { 4, 5, 6, 8, 9, 435, 3454, 345543 };int bitmapA[1 + N / WORD];int bitmapB[1 + N / WORD];int i = 0;int j = 0;for (i = 0; i < N; i++) {clear(bitmapA, i);clear(bitmapB, i);}for (i = sizeof(A) / sizeof(*A) - 1; i >= 0; i--) {set(bitmapA, *(A + i));}for (j = sizeof(B) / sizeof(*B) - 1; j >= 0; j--) {set(bitmapB, *(B + j));}for (i = 0; i < N; i++) {if (test(bitmapA, i) & test(bitmapB, i)) {//交集printf("%d/n", i);}}}int main(void) {//sort();merge();return EXIT_SUCCESS;}