Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined)

B. Code For 1
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 n, l, r (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.
Examples
Input
Copy

7 2 5

Output
Copy

4

Input
Copy

10 3 10

Output
Copy

5

Note

Consider first example:

Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.

For the second example:

Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.

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二分处理；
可以发现每次实际都是对称的；
显然可以求出总的区间长度，然后类似与线段树的操作，每次二分区间，顺便也统计答案即可，orz ；
（思维题）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 500005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
typedef pair<int, int> pii;


inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

ll n;
ll ans;

void dfs(ll l, ll r, ll L, ll R, ll sum) {
	if (l > r || L > R)return;
	ll mid = (l + r) >> 1;
	if (R < mid)dfs(l, mid - 1, L, R, sum / 2);
	else if (L > mid)dfs(mid + 1, r, L, R, sum / 2);
	else {
		ans += sum % 2;
		dfs(l, mid - 1, L, mid - 1, sum / 2);
		dfs(mid + 1, r, mid + 1, R, sum / 2);
	}
}


int main()
{
	//ios::sync_with_stdio(false);
	rdllt(n); ll l, r;
	rdllt(l); rdllt(r);
	ll s = 1;
	ll tmp = n;
	while (tmp>1) {
		tmp /= 2;
		s = s * 2 + 1;
	}
	dfs(1, s, l, r, n);
	cout << ans << endl;
}