Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) B. Code For 1（DFS好题）

B. Code For 1
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
7 2 5
output
4
input
10 3 10
output
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.

题意：对于给定的数字N，可以分解为｛N/2，N%2，N/2｝三个数，接着继续分解两个N/2，直至分解出的三个数全是0或1为止，这样会把N分解为长度为len的01串，接着统计出给定L~R范围内的01串有多少个1。

思路：类似于线段树/二叉树的变形，对于每一个结点都深搜到最底层，再根据L~R的范围和mid以及N/2的对称性返回相应结果。

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;long long int DFS(long long int N,long long int L,long long int R){    if(L>R||N==0)        return 0;    if(N==1)        return 1;    long long int mid=(long long int)(1)<<((int)log2(N));//log以2为底N的对数    //如果采用log(N)/log(2)会有精度误差    if(L>mid)//完全在区间右侧        return DFS(N/2,L-mid,R-mid);//对称结构    if(R<mid)//完全在区间左侧       return DFS(N/2,L,R);    return DFS(N/2,L,mid-1)+DFS(N/2,1,R-mid)+N%2;}int main(){    long long int N,L,M;    cin>>N>>L>>M;    cout<<DFS(N,L,M)<<endl;    return 0;}