[题目链接]
http://poj.org/problem?id=3252
[算法]
数位DP
f[i][j][k]表示在二进制表示下第1位为i,共有j位,其中共有k位为0
实现时需注意细节!
[代码]
#include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #include <deque> #include <exception> #include <fstream> #include <functional> #include <limits> #include <list> #include <map> #include <iomanip> #include <ios> #include <iosfwd> #include <iostream> #include <istream> #include <ostream> #include <queue> #include <set> #include <sstream> #include <stdexcept> #include <streambuf> #include <string> #include <utility> #include <vector> #include <cwchar> #include <cwctype> #include <stack> #include <limits.h> using namespace std; long long S,E; int f[10][50][50]; inline void dp() { int i,j; f[1][1][0] = f[0][1][1] = 1; for (i = 1; i < 31; i++) { for (j = 0; j <= i; j++) { f[0][i + 1][j + 1] += f[0][i][j] + f[1][i][j]; f[1][i + 1][j] += f[0][i][j] + f[1][i][j]; } } } inline int calc(int x) { int i,j,len = 0,cnt = 0; int a[50]; int ret = 0; memset(a,0,sizeof(a)); while (x) { a[++len] = x & 1; x >>= 1; } reverse(a + 1,a + len + 1); for (i = 1; i < len; i++) { for (j = (i + 1) / 2; j <= i; j++) { ret += f[1][i][j]; } } for (i = 2; i <= len; i++) { if (a[i] == 0) cnt++; if (a[i] == 1) { for (j = (len + 1) / 2; j <= len; j++) if (j >= cnt) ret += f[0][len - i + 1][j - cnt]; } } return ret; } int main() { dp(); scanf("%d%d",&S,&E); printf("%d ",calc(E + 1) - calc(S)); return 0; }