Codeforces Good Bye B. New year and old property

Title Link: Http://codeforces.com/problemset/problem/611/B

The title means: in [A, b] this range (1≤ a ≤ b ≤10^)Statistics of the number of matching binary means that only 1 0 of the quantity is what.

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First post a piece of code that will definitely time out, even the courage to submit the wood has it ~ because the fourth group of data is approaching the turtle speed, so ... (Everyone can ignore)

　　

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5 using namespacestd;6 7 typedef __int64 LL;8 ll A, B;9 ll ans;Ten  One intMain () A { - #ifndef Online_judge -Freopen ("In.txt","R", stdin); the     #endif //Online_judge -  -      while(SCANF ("%i64d%i64d", &a, &b)! =EOF) { - __int64 i; +Ans =0; -  +          for(i = A; I <= B; i++) { A__int64 L =i; at            intc =0; -             while(l && C <=1) { -c + = ((L &1) ==0?1:0); -L >>=1; -           } -           if(c = =1) inans++; -  to         } +printf"%i64d\n", ans); -     } the     return 0; *}

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Talk about the right way to solve the problem-that is, the violent enumeration .... Of course, not the decimal number first into a binary number and then slowly compared to, I above the code is similar to this thought, said many are tears, Woo Woo ~ ~ ~

is the analog binary number, in the [1, 10^18] range, the statistics show only binary representation of 0 occurrences 1 times

For the first time, I knew it was possible. = = 、、、 slowly accumulate experience ~ ~ ~

In addition to say a number of the range, 2^61 is already bigger than 10^18, so when enumerating, for example to 61 times on the line, the code has detailed comments, we slowly appreciate ^_^

　　

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5 using namespacestd;6 7 typedef __int64 LL;8 Const intbit = A;//2^61 = 1,152,921,504,606,846,9769 Ten intMain () One { A #ifndef Online_judge -Freopen ("In.txt","R", stdin); -     #endif //Online_judge the  - ll A, B; -      while(SCANF ("%i64d%i64d", &a, &b)! =EOF) { -         intAns =0; +  -          for(inti =2; I <= bit; i++) { +ll mea = (1ll << i)-1;//1, one, 111, 1111, ... A  at              for(intj =0; J < i-1; J + +) { -ll t = mea-(1ll << J);//(1LL<<J): 1, ten , ... -Ans + = (t >= a && t <=b); -             } -         } -printf"%d\n", ans); in     } -     return 0; to}

　　

　　

Codeforces Good Bye B. New year and old property