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 {//数字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 }
题目:
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:
非常有趣的一道题,就是说我们有一个数n我们把它打散为一串0,1序列,方式如上所述,问l到r中有多少个1。
打散方式:比如说7,。
先通过找规律我们可以发现以下两条结论:
1,数x打散后的序列中有x个1。
2,数x打散后的序列有 2n-1位数,其中2n-1为刚好大于等于x的数。
现在给你l,r,问从第l位到第r位有多少个1.
那么这个问题,我们就可以转化为第x位以前有多少个1,然后再相减,就是答案了。
我们可以采用线段树的思想,
这样我们就可以利用第2条确定向左走还是向右走,第1条来确定有多少个1了。
http://codeforces.com/problemset/problem/768/B B. Code For 1
原文:http://www.cnblogs.com/sun-yinkai/p/7826108.html