owb form target pla rst phy ace empty creat

題目鏈接

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 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. 題意：每次查詢有兩種操作 op1：求加入L~R的數時所消耗的單元 op2：求將x加入集合或移動到其它集合所消耗的單元（即由x引起消耗的單元） 思路：op1：每次加入一個數i 那麽會移動[i-lowbit(i)+1 , i-1] ，總的消耗是i-(i-lowbit(i)+1) +1=lowbit(i) 所以每次加入一個數對應的消耗是2的冪次，那麽求L~R即可以枚舉冪次，即： ans+=(n/(1<<i)-n/(1<<(i+1)))*(1<<i) 解釋一下，n/(1<<i)-n/(1<<(i+1))表示長為2^i的消耗的數的個數，例如：n=10 ， 包含長為2的數是2，6，10 為什麽4，8不是，因為它們雖然是2的倍數，但更是4的倍數，包含更長的區間了，所以這部分要減去。 op2：由樹狀數組可知 [i-lowbit(i)+1 , i-1] 是以i為根節點對應的區間，如果假如的數能夠移動i ，那麽這個數對應的孩子區間一定包含i ，所以從x向上一直找父節點即可。 代碼如下：

#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=0;
    while(x<=n)
    {
        ans++;
        x+=lowbit(x);
    }
    return ans;
}
LL cal(LL x)
{
    LL ans=0;
    LL tmp=1;
    for(LL i=0; tmp<=x; i++)
        ans+=(x/(tmp)-x/(tmp<<1))*tmp,tmp<<=1;
    return ans;
}
int main()
{
    LL n,q;
    while(scanf("%lld%lld",&n,&q)!=EOF)
    {
        while(q--)
        {
            int op;
            scanf("%d",&op);
            if(op==1)
            {
                LL x,y;
                scanf("%lld%lld",&x,&y);
                LL ans=cal(y)-cal(x-1);
                printf("%lld\n",ans);
            }
            else
            {
                LL x;
                scanf("%lld",&x);
                LL ans=query(x,n);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}

hdu 5975---Aninteresting game（樹狀數組）