SBT解决约瑟夫环问题（poj3517）

And Then There Was One

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 4852 Accepted: 2581

Description

Let’s play a stone removing game.

Initially, n stones are arranged on a circle and numbered 1, …, n clockwise (Figure 1). You are also given two numbersk and m. From this state, remove stones one by one following the rules explained below, until only one remains. In step 1, remove stonem. In step 2, locate the k-th next stone clockwise from m and remove it. In subsequent steps, start from the slot of the stone removed in the last step, makek hops clockwise on the remaining stones and remove the one you reach. In other words, skip (k − 1) remaining stones clockwise and remove the next one. Repeat this until only one stone is left and answer its number. For example, the answer for the case n = 8, k = 5, m = 3 is 1, as shown in Figure 1.

Initial state
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Final state Figure 1: An example game

Initial state: Eight stones are arranged on a circle.

Step 1: Stone 3 is removed sincem = 3.

Step 2: You start from the slot that was occupied by stone 3. You skip four stones 4, 5, 6 and 7 (sincek = 5), and remove the next one, which is 8.

Step 3: You skip stones 1, 2, 4 and 5, and thus remove 6. Note that you only count stones that are still on the circle and ignore those already removed. Stone 3 is ignored in this case.

Steps 4–7: You continue until only one stone is left. Notice that in later steps when only a few stones remain, the same stone may be skipped multiple times. For example, stones 1 and 4 are skipped twice in step 7.

Final State: Finally, only one stone, 1, is on the circle. This is the final state, so the answer is 1.

Input

The input consists of multiple datasets each of which is formatted as follows.

n k m

The last dataset is followed by a line containing three zeros. Numbers in a line are separated by a single space. A dataset satisfies the following conditions.

2 ≤ n ≤ 10000, 1 ≤ k ≤ 10000, 1 ≤ m ≤ n

The number of datasets is less than 100.

Output

For each dataset, output a line containing the stone number left in the final state. No extra characters such as spaces should appear in the output.

Sample Input

8 5 3100 9999 9810000 10000 100000 0 0

Sample Output

1932019

Source

Japan 2007

跟上一个题一样的，只不过是是输出最后一个

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;struct SBT{    int left,right,size,key;    void init(int val)    {        left=right=0;        size=1;        key=val;    }}tree[maxn];int tot,root;void left_rotate(int &x){    int y=tree[x].right;    tree[x].right=tree[y].left;    tree[y].left=x;    tree[y].size=tree[x].size;    tree[x].size=tree[tree[x].left].size+tree[tree[x].right].size+1;    x=y;}void right_rotate(int &x){    int y=tree[x].left;    tree[x].left=tree[y].right;    tree[y].right=x;    tree[y].size=tree[x].size;    tree[x].size=tree[tree[x].left].size+tree[tree[x].right].size+1;    x=y;}void maintain(int &x,int flag){    if(!flag)    {        if(tree[tree[tree[x].left].left].size>tree[tree[x].right].size)            right_rotate(x);        else if(tree[tree[tree[x].left].right].size>tree[tree[x].right].size)            left_rotate(tree[x].left),right_rotate(x);        else return;    }    else    {        if(tree[tree[tree[x].right].right].size>tree[tree[x].left].size)            left_rotate(x);        else if(tree[tree[tree[x].right].left].size>tree[tree[x].left].size)            right_rotate(tree[x].right),left_rotate(x);        else return;    }    maintain(tree[x].left,0);    maintain(tree[x].right,1);    maintain(x,0);    maintain(x,1);}//插入值为key的节点void insert(int &x,int key){    if(!x)    {        x=++tot;        tree[x].init(key);    }    else    {        tree[x].size++;        if(key<tree[x].key)insert(tree[x].left,key);        else insert(tree[x].right,key);        maintain(x,key>=tree[x].key);    }}//删除值为key的节点int del(int &x,int key){    if(!x)return 0;    tree[x].size--;    if(key==tree[x].key||(key<tree[x].key&&tree[x].left==0)||       (key>tree[x].key&&tree[x].right==0))    {        if(tree[x].left&&tree[x].right)        {            int p=del(tree[x].left,key+1);            tree[x].key=tree[p].key;            return p;        }        else        {            int p=x;            x=tree[x].left+tree[x].right;            return p;        }    }    else        return del(key<tree[x].key?tree[x].left:tree[x].right,key);}int get_kth(int x,int k){    if(k<=tree[tree[x].left].size)        return get_kth(tree[x].left,k);    else if(k>tree[tree[x].left].size+1)        return get_kth(tree[x].right,k-tree[tree[x].left].size-1);    return tree[x].key;}int N,K,M;int main(){      while(scanf("%d%d%d",&N,&K,&M)!=EOF,N||M||K)    {        root=tot=0;        for(int i=1;i<=N;i++)insert(root,i);        bool first=true;        int last=M+1-K;        for(int i=1;i<N;i++)        {            int k=(K+last-1)%tree[root].size;            if(!k)k=tree[root].size;            int tmp=get_kth(root,k);            del(root,tmp);            last=k;        }        printf("%d\n",tree[root].key);    }    return 0;}