Codeforces 768B. Code For 1

Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility. 

Initially Sam has a list with a single element n. Then he has to perform certain operations on this list. In each operation Sam must remove any element x, such that x > 1, from the list and insert at the same position  sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.

Now the masters want the total number of 1s in the range l to r (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?

Input

The first line contains three integers nlr (0 ≤ n < 250, 0 ≤ r - l ≤ 105, r ≥ 1, l ≥ 1) – initial element and the range l to r.

It is guaranteed that r is not greater than the length of the final list.

Output

Output the total number of 1s in the range l to r in the final sequence.

 分治法解决,主要是处理边界很难受
#include <cstdio>
#include <cmath>
#include <cctype>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <stack>
#include <vector>
#include <map>
#include <set>
using namespace std;
typedef long long LL;

LL now = 0,n,l,r;
LL sq[10000000] = {0};
void divide(LL n){
  if (n == 1 || n == 0){
    now ++;
    sq[now] = n; return;
  }
  divide(n/2);
  int nowi = now;
  divide(n%2);
  for (int i=now+1;i<=now+nowi;i++){
    sq[i] = sq[i-now];
  }
  now = now + nowi;
}

LL f(LL n){
  if (n == 1 || n == 0){
    return 1;
  }
  else return 2*f(n/2) + f(n % 2);
}

LL calc(LL l,LL r,LL n){
  LL half = f(n/2);
  if (l > r) return 0;
  if (n == 1){
    return 1;
  }
  if (r <= half){
    return calc(l,r,n/2);
  }

  // r > half
  if (r == half + 1){
    return calc(l,half,n/2) + (n % 2);
  }
  // r > half + 1
  if (l > half){
    if (l == half + 1){
      return calc(1,r-half-1,n/2) + (n % 2);
    }
    else {
      return calc(l-half-1,r-half-1,n/2);
    }
  }
  else { // l <= half
    return calc(l,half,n/2) + calc(1,r-half-1,n/2) + (n % 2);
  }
}

int main() {
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);
  cin >> n >> l >> r;
  if (n == 0) {
    cout << 0; return 0;
  }
  cout << calc(l,r,n);


  // LL ans = 0;
  // for (int i=l;i<=r;i++){
  //   if (sq[i] == 1){
  //     ans ++;
  //   }
  // }
  // cout << ans;
  // for (int i=1;i<=now;i++){
  //   cout << sq[i];
  // }
  return 0;
}
View Code
原文地址:https://www.cnblogs.com/ToTOrz/p/7275638.html