poj3252 Round Numbers

poj3252 Round Numbers

Combinatorial Math (digital DP)

(I did not expect to see the mark in the book to face Tat)

(I will no longer use scanf/printf tat on POJ)

(POJ's title is the default for multiple sets of data, TAT)

Tips: Information Science Orsay Mathematics a pass of the standard and part of the code on the Baidu Access array out of bounds , face Black will GG (such as I qaq), detailed see

Test instructions: The number of numbers in the closed interval [A, a] is converted to binary after 0 is more than 1, 1<=a<b<=2* (10^9)

is obviously a digital DP routine: Solve (b+1)-solve (a) (the Solve (n) of the subject does not include N)

So the scope of our settlement becomes [0,n].

First turn n into binary (gray often obviously), throw into the array wt

Blue after we sort of discussion

1. Length of the <n length of the number to be evaluated

We can calculate the number of combinations directly.

Enumerates the lengths of the I=1 to Len-1-1 (the first bit must be 1 so one more button)

Re-enumeration of the number of 0 j= i/2+1 to I, accumulate C (i, j) it's over.

2. Length of the ==n length of the number to be evaluated

We take the WT array from high to low enumeration i= len-1 to 1 (∵↑↑∴len-1)

When wt[i]==0, we use the variable Z to count 0

When Wt[i]==1:

Let's first assume that this bit is 0, then the next few bits are strictly less than n, regardless of the rank

∴ can also enumerate the number of 0 in combination number blind

J= Max (0, (len+1)/2-(z+1)) to I-1

attention: Take Max to prevent the (negative) bounds of the array (the standard pot is here)

Finally, the answer is summed up, end.

Use 2h30min,6 submissions TAT

#include <iostream>using namespacestd;intMaxintAintb) {returnA>b?a:b;}intwt[ *],a,b,c[ *][ *];intSolveintN) {    intres=0, z=0, len=0;  for(; n;n>>=1) wt[++len]= (n&1); Turn Binary for(intI=1; i<len-1;++i)//length <n for(intj=i/2+1; j<=i;++j) Res+=C[i][j];  for(inti=len-1; i>=1;--i)//length ==nif(Wt[i]) { for(intJ=max (0, (len+1)/2-z-1); j<i;++j) Res+=c[i-1][j]; }Else++Z; returnRes;}intMain () { for(intI=0; i<= +;++i) for(intj=0; j<=i;++j) C[i][j]= (!j| | I==J)?1: c[i-1][j-1]+c[i-1][j];  while(cin>>a>>b) Cout<<solve (b +1)-solve (a) <<Endl; return 0;}

poj3252 Round Numbers