hdu 5860 Death Sequence（2016 Multi-University Training Contest 10——递推）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5860

Death Sequence

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 404    Accepted Submission(s): 179

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

17 2 71234567

 

Sample Output

1357264

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

题目大意：给你n个人排成一列编号，每次杀第一个人第i×k+1个人一直杀到没的杀。然后剩下的人重新编号从1～剩余的人数。按照上面的方式杀。问第几次杀的是谁。

解题思路：这道题首先我们先将编号改成从0开始。这样如果i%k==0那么我们可以知道第i个数字第一轮就被杀死。如果i%k!=0，那么我们知道在这之前一轮杀死了i/k+1个人。这样我们就得到递推式。

详见代码。

//输出第几个被杀的是谁#include <iostream>#include <cstdio>#include <algorithm>using namespace std;#define N 3000000+10struct node{    int d;//表示哪一轮被杀的    int num;//表示当前轮第几个被杀的    int p;//表示本身是几号}s[N];bool cmp(node a,node b){    if (a.d==b.d)        return a.num<b.num;    else        return a.d<b.d;}int main(){    int t,a;    scanf("%d",&t);    while (t--)    {        int n,k,q;        scanf("%d%d%d",&n,&k,&q);        s[0].num=1;        for (int i=0;i<n;i++)        {            s[i].p=i+1;            if (i%k==0)            {                s[i].d=1;                if (i==0)                    continue;                s[i].num=s[i-k].num+1;            }            else            {                s[i].d=s[i-(i/k+1)].d+1;                s[i].num=s[i-(i/k+1)].num;            }        }          sort(s,s+n,cmp);        for (int i=0;i<q;i++)        {            scanf("%d",&a);            printf ("%d\n",s[a-1].p);        }    }    return 0;}