CF 431 D. Random Task Combination Math

Test instructions

Given M,k

0 <= M <= 10^18, 1 <= k <= 64

Find a number n, satisfy the n+1,n+2,... n+n This n number, just has the m number of 2 binary notation just have K 1

Ensure that the answer is within 10^18

Ideas:

Obviously

For x, if x+1,x+2,..., x+x has a y number of K 1

For X+1, the x+2,x+3,..., x+x+2 has a number of K 1 and >= y

To satisfy Monotonicity, consider two points:

L = M,r = 10^18

So the question becomes: given a number x,x+1,x+2,..., the number of K-1 in X+x is exactly how many

For ease of expression, assume that X is a 2 binary number:

Then the range of K 1 appears:

[X,11...111],[10...000,2X]

Which is divided into 2 segments according to the number of digits.

function Go (LL x,int k) function: For [x,1111111] This interval has K 1 number of how many

Then ans = Go (x,k) + Go (10...000,k)-Go ((x<<1) &1,k)

Judging the relationship between ANS and M can reduce the range of l,r.

Attention:

For 2^x (1 << x), this time (1 << x) is the default int type,

So if it's super int, use (1LL << x)

Code:

                                              //File Name:cf431D.cpp//Author:long//Mail: [email protected]//Created time:2016 July 09 Saturday 21:34 35 seconds#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#defineLL Long Longusing namespacestd;inta[ -],tot; LL f[ -][ -];voidinit () {memset (F,0,sizeoff);  for(intI=0;i< -; i++) {f[i][0] =1;  for(intj=1; j<=i;j++) F[i][j]= f[i-1][J] + f[i-1][j-1]; }}ll Go (LL x,intk) {Tot=0;  while(x) {a[++tot] = x%2; X>>=1; } LL ans=0; intPre =0;  for(intI=tot;i>0; i--){        if(A[i] = =1) Pre++; Else{            if(K-pre-1>=0) ans+ = f[i-1][k-pre-1]; Else                  Break; }    }    if(Pre = =k) Ans++; returnans;} LLGet(LL x,intk) {LL ans=Go (x,k); LL y= (1LL <<tot); //printf ("x =%lld y =%lld\n", x, y);Ans = ans + go (y,k)-Go (2* x +1, K); returnans;} ll solve (ll M,intk) {LL L= M,r = (LL) 1e18 +1, Mid;  while(R-l >1) {Mid= (L + r) >>1; LL cur=Get(MID,K); //printf ("mid =%lld cur =%lld\n", mid,cur);        if(cur <=m) L=mid; ElseR=mid; }    if(Get(l,k) = =m)returnl; Else         returnR;}intMain () {init ();    LL m; intK; CIN>> m >>K; cout<< Solve (m,k) <<Endl; return 0;}

CF 431 D. Random Task Combination Math