HDU 5860 Death Sequence (递推)

题意：n个人排成一行，从第一个人开始，每个k个人报数，报到数的人被杀死，剩下的人重新排成一行再报数。一共q个询问，每次询问第qi个死的人是谁。

析：是一个约瑟夫的变形，我们要考虑子问题的问题同样编号是0-n-1，如果在某一轮，第 i 个人如果能取模 k 为0，那么这一轮他就会被干掉，如果不是

那么下一轮他就是编号为 i-i/k-1，然后再处理每一轮多少个人被干掉，就OK了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3000000 + 10;
const int mod = 1000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn], sum[maxn], f[maxn], t[maxn], ans[maxn];

int main(){
  int T;  cin >> T;
  while(T--){
    int q;
    scanf("%d %d %d", &n, &m, &q);
    int nn = n, cnt = 1;
    while(nn){
      sum[cnt] = sum[cnt-1];
      sum[cnt++] += (nn-1) / m + 1;
      nn -= (nn-1) / m + 1;
    }

    memset(t, 0, sizeof t);
    for(int i = 0; i < n; ++i){
      dp[i] = i % m ? dp[i-i/m-1] + 1 : 1;
      f[i] = ++t[dp[i]];
    }
    for(int i = 0; i < n; ++i)  ans[sum[dp[i]-1]+f[i]] = i;
    while(q--){
      scanf("%d", &m);
      printf("%d
", ans[m]+1);
    }
  }
  return 0;
}