一、相互转换
注意由于bitset<Size> bs(num);中的Size必须为常数,所以去前导0比较麻烦。
#include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<string> #include<bitset> using namespace std; const int Size = 32; char str[Size]; int main(void) { int num; while (~scanf("%d", &num)) ///禁止输入负值! { ///化成无符号2进制 bitset<Size> bs(num); strcpy(str, bs.to_string().c_str()); ///注意算log要加上1e-9 for (int i = Size - (int)log2(num + 1e-9) - 1; i < Size; ++i) putchar(str[i]); putchar('\n'); ///变回来方法1 string str2(str); bitset<32> bs2(str2); printf("%d\n", bs2.to_ulong()); ///或者写%lud ///变回来方法2,#include<cstdlib> printf("%d\n", strtol(str, NULL, 2)); ///或者写%ld } return 0; }
但悲剧的是bitset只能接收32位的无符号整数,所以我们需要自己写一个:
#include<cstdio> #include<cstring> #include<iostream> using namespace std; typedef unsigned long long ull; const int mx = 65; char bits[mx], tmp[mx]; void getBitsWithPreZero(ull n, int len) { int i = 0; while (n) { tmp[i++] = '0' + (n & 1ULL); n >>= 1ULL; } memset(bits, '0', sizeof(bits)); for (int j = len - i; j < len; ++j) bits[j] = tmp[--i]; bits[len] = 0; } void getBitsWithoutPreZero(ull n) { int i = 0, j = 0; while (n) { tmp[i++] = '0' + (n & 1ULL); n >>= 1ULL; } while (i) bits[j++] = tmp[--i]; bits[j] = 0; } int main() { ull n; int len; while (cin >> n >> len) { getBitsWithPreZero(n, len); puts(bits); getBitsWithoutPreZero(n); puts(bits); } return 0; }
二、生成0~2^k的二进制数
无前导零:
#include<cstdio> #include<string> using namespace std; const int maxn = 1 << 7; string bs; int main(void) { int i, ii; ///无前导零 for (i = 0; i < maxn; ++i) { ii = i; bs = ""; do { bs = (ii & 1 ? "1" : "0") + bs; ii >>= 1; } while (ii); puts(bs.c_str()); } return 0; }
有前导零:
#include<cstdio> #include<string> #include<bitset> using namespace std; const int Size = 4; const int maxn = 1 << Size; int main(void) { for (int i = 0; i < maxn; ++i) { bitset<Size> bs(i); puts(bs.to_string().c_str()); } return 0; }