[jzoj1408][vijos1472] 教主的集合序列

vijos 或 jzoj

Solution

显然的，每次得到的集合都会是区间
根据条件，每次区间会扩大一倍
第i个区间来讲
左端点为 2i−1
右端点为 2i∗n−2i−1+1
因而可以依次枚举每个区间的左右端点 加入高精度即可

其实，有一种更优秀的方式
可以发现，每个区间的大小可以被表达为

(n−3)∗2i−1+3

那么 前i个区间元素个数和也可以被计算
这就满足二分了

Code

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define oo 2139062143#define sqr(x) ((x)*(x))#define lowbit(x) ((x)&(-x))#define abs(x) (((x)>=0)?(x):(-(x)))#define max(x,y) (((x)>(y))?(x):(y))#define min(x,y) (((x)<(y))?(x):(y))#define fo(i,x,y) for (int i = (x);i <= (y);++ i)#define fd(i,x,y) for (int i = (x);i >= (y);-- i)using namespace std;typedef double db;typedef long long ll;const int M = 8080, JW = 10000000,N = 1010;ll n,m,ans[N][M];  ll k[M];  void write(ll a[])  {      printf("%lld",a[a[0]]);      fd(i,a[0] - 1,1) printf("%07lld",a[i]);  }  int cmp(ll a[],ll b[])//-1:a=b  0:a<b  1:a>b  {        if (a[0] != b[0]) return (a[0] > b[0]);      fd(i,a[0],1)    if (a[i] != b[i]) return (a[i] > b[i]);      return -1;}  void mult(ll a[],ll b[],ll c[])  {      ll t[M]= {0};      t[0] = a[0] + b[0];      fo(i,1,a[0]) fo(j,1,b[0])        t[i + j - 1] += a[i] * b[j];    fo(i,1,t[0]) t[i + 1] += t[i] / JW,t[i] %= JW;    while (!t[t[0]] && t[0] > 1) -- t[0];      memcpy(c,t,sizeof(t));  }  void qsm(ll a[],ll x)  {      ll t[M]= {0};    t[0] = 1,t[1] = 2;    a[0] = a[1] = 1;      while (x)      {          if (x & 1) mult(a,t,a);          mult(t,t,t);          x >>= 1;      }  }  void ins(char s[],ll a[])  {      int l = strlen(s);    a[0] = 1;    int tmp;    for (tmp = l;tmp > 6;tmp -= 7)    {          fo(j,0,6) a[a[0]] = a[a[0]] * 10 + (s[tmp - 7 + j] -'0');        ++ a[0];     }      for (int j = 0;j < tmp;++ j)        a[a[0]] = a[a[0]] * 10 + s[j] - '0';}  void summing(ll a[],ll b[],ll c[])  {      ll t[M] = {0};      t[0]=max(a[0],b[0]);      fo(i,1,t[0]) t[i] = a[i] + b[i];    fo(i,1,t[0]) t[i + 1] += t[i] / JW,t[i] %= JW;    while (t[t[0] + 1]) ++ t[0];      memcpy(c,t,sizeof(t));  }  void minu(ll a[],ll b[],ll c[])  {      ll t[M]= {0};    t[0] = a[0];      fo(i,1,t[0]) t[i] = a[i] - b[i];    fo(i,1,t[0] - 1)     if (t[i] < 0) -- t[i + 1],t[i] += JW;    while (!t[t[0]] && t[0] > 1) -- t[0];      memcpy(c,t,sizeof(t));  }  void divt(ll a[])  {      fd(i,a[0],2)    {          if (a[i] & 1) a[i - 1] += JW;          a[i] >>= 1;      }      a[1] >>= 1;      while (!a[a[0]] && a[0] > 1) -- a[0];  }  void sigc(ll t[],ll mid,ll tt[])  {      qsm(t,mid - 1);      tt[1] = n - 3;      mult(t,tt,t);      tt[1] = mid * 3;      summing(t,tt,t);      tt[1] = n;      minu(t,tt,t);  }  void work()  {      ll t[M]= {0},tt[M]= {0};    tt[0] = 1;     int l = 1,r = 5000;      while (l < r)      {          int mid = (l + r + 1) >> 1;          sigc(t,mid,tt);          int j = cmp(k,t);          if (j == -1)          {              qsm(t,mid-2);              tt[1] = n - 1;              mult(t,tt,t);              tt[1] = 1;              summing(t,tt,t);              write(t);              return;          }          if (j) l = mid;          else r = mid - 1;      }      sigc(t,r,tt);      minu(k,t,k);      qsm(t,r);      tt[1] = 2;      minu(t,tt,t);      summing(k,t,t);      write(t);  }  int main()  {      ans[1][1] = 1;    ans[2][1] = 1,ans[2][2] = 2,ans[2][3] = 3;    char ch[N];    scanf("%lld", &n);        if (n < 4)          {              scanf("%lld",&m);            if (n <= 2)                 if (ans[n][m] == 0) printf("-1");                else printf("%lld", ans[n][m]);            if (n == 3)              {                  qsm(k,(m + 2) / 3);                  if (m % 3 == 1) -- k[1];                  if (m % 3 == 0) ++ k[1];                  write(k);              }            return 0;        }          memset(k,0,sizeof k);          scanf("%s", ch);          ins(ch,k);          work();      printf("\n");}