HDU 5860 Death Sequence （递推 + 约瑟夫环 + 思维）——2016 Multi-University Training Contest 10

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Death Sequence

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 379    Accepted Submission(s): 163

Problem Description

You may heard of the Joseph Problem, the story comes from a Jewish historian living in 1st century. He and his 40 comrade soldiers were trapped in a cave, the exit of which was blocked by Romans. They chose suicide over capture and decided that they would form a circle and start killing themselves using a step of three. Josephus states that by luck or maybe by the hand of God, he and another man remained the last and gave up to the Romans.

Now the problem is much easier: we have N men stand in a line and labeled from 1 to N, for each round, we choose the first man, the k+1-th one, the 2*k+1-th one and so on, until the end of the line. These poor guys will be kicked out of the line and we will execute them immediately (may be head chop, or just shoot them, whatever), and then we start the next round with the remaining guys. The little difference between the Romans and us is, in our version of story, NO ONE SURVIVES. Your goal is to find out the death sequence of the man.

For example, we have N = 7 prisoners, and we decided to kill every k=2 people in the line. At the beginning, the line looks like this:

1 2 3 4 5 6 7

after the first round, 1 3 5 7 will be executed, we have

2 4 6

and then, we will kill 2 6 in the second round. At last 4 will be executed. So, you need to output 1 3 5 7 2 6 4. Easy, right?

But the output maybe too large, we will give you Q queries, each one contains a number m, you need to tell me the m-th number in the death sequence.

 

Input

Multiple cases. The first line contains a number T, means the number of test case. For every case, there will be three integers N (1<=N<=3000000), K(1<=K), and Q(1<=Q<=1000000), which indicate the number of prisoners, the step length of killing, and the number of query. Next Q lines, each line contains one number m(1<=m<=n).

 

Output

For each query m, output the m-th number in the death sequence.

 

Sample Input

1 
7 2 7 
1 
2 
3 
4 
5 
6 
7

 

Sample Output

1 
3 
5 
7 
2 
6 
4

题目大意：

有 n 个人排成一列编号从 1−n，每次杀第一个人，然后是从第 i∗k+1 个人一直杀到最后，然

后剩下的人在重新编号从 1～x。按照上面的方式继续杀。问第 i 次杀的是谁，输出序号。

解题思路：

首先将 n 个人从 0 开始编号，然后定义一个 pair<int,int> 分别表示：

first：i 是在第 p[i].first轮 死的

second: i 是在第 p[i].first轮 第 p[i].second 死的

然后第一轮如果满足 i MOD k==0 的话，那么 p[i].first=0,p[i].second=i/k+1 否

则的话，根据前面的 i/k+1 的关系满足一下公式：

p[i].first = p[i−i/k−1].first+1;

p[i].second = p[i−i/k−1].second;

然后再用一个 sum 数组，sum[i] 表示 第 i 轮 一共死多少人，最后统计一下就行了。

/**2016 - 08 - 18 下午Author: ITAKMotto:今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。**/#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;typedef long long LL;typedef unsigned long long ULL;const int INF = 1e9+5;const int MAXN = 3e6+5;const int MOD = 1e9+7;const double eps = 1e-7;const double PI = acos(-1);using namespace std;LL Scan_LL()///输入外挂{    LL res=0,ch,flag=0;    if((ch=getchar())=='-')        flag=1;    else if(ch>='0'&&ch<='9')        res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        res=res*10+ch-'0';    return flag?-res:res;}int Scan_Int()///输入外挂{    int res=0,ch,flag=0;    if((ch=getchar())=='-')        flag=1;    else if(ch>='0'&&ch<='9')        res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        res=res*10+ch-'0';    return flag?-res:res;}void Out(LL a)///输出外挂{    if(a>9)        Out(a/10);    putchar(a%10+'0');}pair <int,int> p[MAXN];/**first：i 是在第 p[i].first轮 死的second: i 是在第 p[i].first轮 第 p[i].second 死的**/int sum[MAXN], ans[MAXN];/**sum: 第 i 轮 一共死多少人**/void Init(int n, int k){    for(int i=0; i<n; i++)    {        if(i%k == 0)        {            p[i].first = 0;            p[i].second = i/k+1;            ///cout<<"i = "<<i<<"   first = "<<p[i].first<<"  second = "<<p[i].second<<endl;        }        else        {            p[i].first = p[i-i/k-1].first+1;            p[i].second = p[i-i/k-1].second;            ///cout<<"i = "<<i<<"   first = "<<p[i].first<<"  second = "<<p[i].second<<endl;        }    }    int tmp = n-1, cnt = 0;    sum[0] = 0;    while(tmp)    {        cnt++;        sum[cnt] = sum[cnt-1] + tmp/k+1;        tmp = tmp-(tmp-1)/k-1;    }    for(int i=0; i<n; i++)        ans[sum[p[i].first]+p[i].second] = i+1;}int main(){    int T, n, k, q;    T = Scan_Int();    while(T--)    {        n = Scan_Int();        k = Scan_Int();        q = Scan_Int();        Init(n, k);        while(q--)        {            int x;            x = Scan_Int();            printf("%d\n",ans[x]);        }    }    return 0;}