Problem Description
Let’s play a game.We add numbers 1,2...n in increasing order from 1 and put them into some sets.
When we add i,we must create a new set, and put iinto it.And meanwhile we have to bring [i-lowbit(i)+1,i-1] from their original sets, and put them into the new set,too.When we put one integer into a set,it costs us one unit physical strength. But bringing integer from old set does not cost any physical strength.
After we add 1,2...n,we have q queries now.There are two different kinds of query:
1 L R:query the cost of strength after we add all of [L,R](1≤L≤R≤n)
2 x:query the units of strength we cost for putting x(1≤x≤n) into some sets.
When we add i,we must create a new set, and put iinto it.And meanwhile we have to bring [i-lowbit(i)+1,i-1] from their original sets, and put them into the new set,too.When we put one integer into a set,it costs us one unit physical strength. But bringing integer from old set does not cost any physical strength.
After we add 1,2...n,we have q queries now.There are two different kinds of query:
1 L R:query the cost of strength after we add all of [L,R](1≤L≤R≤n)
2 x:query the units of strength we cost for putting x(1≤x≤n) into some sets.
Input
There are several cases,process till end of the input.
For each case,the first line contains two integers n and q.Then q lines follow.Each line contains one query.The form of query has been shown above.
n≤10^18,q≤10^5
For each case,the first line contains two integers n and q.Then q lines follow.Each line contains one query.The form of query has been shown above.
n≤10^18,q≤10^5
Output
For each query, please output one line containing your answer for this query
Sample Input
10 2
1 8 9
2 6
Sample Output
9
2
Hint
lowbit(i) =i&(-i).It means the size of the lowest nonzero bits in binary of i. For example, 610=1102, lowbit(6) =102= 210
When we add 8,we should bring [1,7] and 8 into new set.
When we add 9,we should bring [9,8] (empty) and 9 into new set.
So the first answer is 8+1=9.
When we add 6 and 8,we should put 6 into new sets.
So the second answer is 2.
题意：每次查询有两种操作
op1：求加入L~R的数时所消耗的单元
op2：求将x加入集合或移动到其它集合所消耗的单元（即由x引起消耗的单元）
思路：op1：每次加入一个数i 那么会移动[i-lowbit(i)+1 , i-1] ，总的消耗是i-(i-lowbit(i)+1) +1=lowbit(i) 所以每次加入一个数对应的消耗是2的幂次，那么求L~R即可以枚举幂次，即： ans+=(n/(1<<i)-n/(1<<(i+1)))*(1<<i)
解释一下，n/(1<<i)-n/(1<<(i+1))表示长为2^i的消耗的数的个数，例如：n=10 ， 包含长为2的数是2，6，10 为什么4，8不是，因为它们虽然是2的倍数，但更是4的倍数，包含更长的区间了，所以这部分要减去。
op2：由树状数组可知 [i-lowbit(i)+1 , i-1] 是以i为根节点对应的区间，如果假如的数能够移动i ，那么这个数对应的孩子区间一定包含i ，所以从x向上一直找父节点即可。
代码如下：
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
LL lowbit(LL x)
{
return x&(-x);
}
LL query(LL x,LL n)
{
LL ans=;
while(x<=n)
{
ans++;
x+=lowbit(x);
}
return ans;
}
LL cal(LL x)
{
LL ans=;
LL tmp=;
for(LL i=; tmp<=x; i++)
ans+=(x/(tmp)-x/(tmp<<))*tmp,tmp<<=;
return ans;
}
int main()
{
LL n,q;
while(scanf("%lld%lld",&n,&q)!=EOF)
{
while(q--)
{
int op;
scanf("%d",&op);
if(op==)
{
LL x,y;
scanf("%lld%lld",&x,&y);
LL ans=cal(y)-cal(x-);
printf("%lld\n",ans);
}
else
{
LL x;
scanf("%lld",&x);
LL ans=query(x,n);
printf("%lld\n",ans);
}
}
}
return ;
}