HDU-5975 Aninteresting game（树状数组原理）

Aninteresting game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 267    Accepted Submission(s): 112

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.

 

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

 

Output

For each query, please output one line containing your answer for this query

 

Sample Input

10 21 8 92 6

 

Sample Output

92
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= 210When 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：从1~n开始插入数字，如果插入x的花费为[x-lowbit(x)+1,x-1]内已插入的数字的数量+1，定义lowbit(x)为x的二进制中为1的最低位

现在给出2种询问：①1 L R：插入[L,R]的花费；②2 x:求x对所有数的贡献

题解：模拟树状数组，具体解析见代码注释

#include<bits/stdc++.h>using namespace std;typedef long long LL;LL f[64];/*****x-1-(x-lowbit(x)+1)+1+1=lowbit(x)，因此每次加入一个数的花费为x的最低位，即是一个2的次幂如果要放入1~n，只要枚举2^k，求1~n中有多少个数的最低位是2^k设最低位是2^k的数有cnt[k]个，f[k]=2^k则cnt[k]=n/f[k]ans+=cnt[k]*f[k]-cnt[k+1]*f[k] 举例解释一下:假设n=10，以2为最低位的是2,6,10,但不包括4,8，因为4,8是以2^2和2^3为最低位的*****/LL solve(LL x){    if(x<=0) return 0;    LL ret=0;    for(int i=0;f[i]<=x;i++) {        LL cnt=x/f[i];        ret+=cnt*f[i];        if(i>0) ret-=cnt*f[i-1];    }    return ret;}int main(){    int m,op;    LL n,l,r,x;    for(int i=0;i<63;i++) f[i]=1ll<<i;    //freopen("in.txt","r",stdin);    while(~scanf("%lld%d",&n,&m)){        while(m--){            scanf("%d",&op);            if(op==1){                scanf("%lld%lld",&l,&r);                printf("%lld\n",solve(r)-solve(l-1));            }            else {                /*****                画个树状数组观察得到：如果y对[x-lowbit(x)+1,x-1]有贡献，那么y的区间一定属于x的区间                例如：3的区间为[3,3],8的区间为[1,8]，那么3是对[1,7]有贡献                因此只要往上找包含y的区间即可                *****/                scanf("%lld",&x);                int ans=0;                while(x<=n){                    ans++;                    x+=x&(-x);                }                printf("%d\n",ans);            }        }    }    return 0;}