1 #include <bits/stdc++.h>
 2 using namespace std;
 3 long long bit[70];
 4 void init()
 5 {
 6     bit[1]=1;
 7     for(int i=2;i<=55;++i)
 8     {
 9         bit[i]=2*bit[i-1]+1;
10     }
11 }
12 long long getRes(long long n,long long  x)
13 {//數字n,第x位
14     if(x==0)return 0;
15     long long
 
 res=0;
16     long long bitsCnt=bit[lower_bound(bit+1,bit+55+1,n/2)-bit];
17     //printf("n=%I64d,x=%I64d,bitsCnt=%I64d\n",n,x,bitsCnt);
18     if(n==1)return 1;
19     if(x==bitsCnt+1)
20     {
21         if(n%2==1)res++;
22         res+=n/2;
23         return res;
24     }
25     else if(x>bitsCnt+1
 
)
26     {
27         if(n%2==1)res++;
28         res+=n/2;
29         res+=getRes(n/2,x-bitsCnt-1);
30     }
31     else 
32     {
33         res+=getRes(n/2,x);
34     }
35     //printf("res=%I64d\n",res);
36     return res;
37 }
38 int main() {
39     long long n,l,r;
40     init();
41     scanf("%I64d%I64d%I64d
 
",&n,&l,&r);
42     if(n==0)
43     {
44         printf("0\n");
45         return 0;
46     }
47     int index=lower_bound(bit+1,bit+55+1,n)-bit;
48     long long res1=getRes(n,l-1);
49     long long res2=getRes(n,r);
50     //printf("res1=%I64d,res2=%I64d\n",res1,res2);
51     printf("%I64d\n",res2-res1);
52     return 0;
53 }

題目：

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?

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 lto r.

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

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

分析：

非常有趣的一道題，就是說我們有一個數n我們把它打散為一串0，1序列，方式如上所述，問l到r中有多少個1。

打散方式：比如說7，。

先通過找規律我們可以發現以下兩條結論:

1，數x打散後的序列中有x個1。

2，數x打散後的序列有 2ⁿ-1位數，其中2ⁿ-1為剛好大於等於x的數。

現在給你l,r，問從第l位到第r位有多少個1.

那麽這個問題，我們就可以轉化為第x位以前有多少個1，然後再相減，就是答案了。

我們可以采用線段樹的思想，

這樣我們就可以利用第2條確定向左走還是向右走，第1條來確定有多少個1了。

http://codeforces.com/problemset/problem/768/B B. Code For 1