POJ 3252 Round Numbers

Title Link: http://poj.org/problem?id=3252

Test instructions: If a binary representation of a number is greater than or equal to 1 of the number of its 0, then this is a round Numbers.

Give an interval and ask how many round Numbers there are in this interval.

Ideas:

Memory search, judge the last number of 0 and 1 of the number, but pay attention to the leading 0 case, the number of leading 0 is removed.

1#include <iostream>2#include <cstring>3#include <cstdio>4 using namespacestd;5 intnum[ +];6 intdp[ +][2][ +][ +];7 intDfsintCurintAllzero,intOneintZerointlimit)8 {9     if(Cur <0)Ten     { One         if(!allzero && (zero >= One))return 1; A         Else return 0; -     } -     if(!limit && Dp[cur][allzero][one][zero]! =-1)returnDp[cur][allzero][one][zero]; the      -     intRET =0; -     intup = Limit?num[cur]:1; -      for(inti =0; I <= up; i++) +     { -         if(i = =0&& Allzero = =0) ret + = DFS (cur-1, Allzero, one, zero+1, limit && i = =Up );  +         Else if(i = =0&& Allzero = =1) ret + = DFS (cur-1, Allzero,0,0, limit && i = =Up ); A         Else if(i = =1&& Allzero = =0) ret + = DFS (cur-1, Allzero, one+1, zero, limit && i = =Up ); at         Else if(i = =1&& Allzero = =1) ret + = DFS (cur-1,0, one+1, zero, limit && i = =Up );  -     } -     if(!limit) Dp[cur][allzero][one][zero] =ret; -     returnret; - } - intSlove (intx) in { -Memset (DP,-1,sizeof(DP)); to     intCNT =0; +      while(x) -     { thenum[cnt++] = x percent2; *X/=2; $     }Panax Notoginseng     returnDFS (cnt-1,1,0,0,1); -      the } + intMain () A { the    //freopen ("In.txt", "R", stdin); +    //freopen ("OUT.txt", "w", stdout); -     intN, M; $      while(~SCANF ("%d%d", &n, &m)) $     { -printf"%d\n", Slove (m)-Slove (n1)); -     } the     return 0; -}

POJ 3252 Round Numbers