AtCoder Regular Contest #043 B問題 - 難易度

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AtCoder Regular Contest #043
問題文

問題概要

$N$ 個の問題があり,$i$ 番目の問題の難易度は $D_i$ である.
この中から $4$ 問を選び,コンテストを行う.
ただし,$k+1$ 問目の難易度は $k$ 問目の難易度の $2$ 倍以上でなければいけない.
これを満たすような問題の選び方は何通りあるかを ${\rm mod}\ 1000000007\ (10^9+7)$ で求める問題.

解法

動的計画法で求める.
状態を($k$ 問目まで選んだ,最後に使った問題)として,そのような選び方のパターン数を計算する.
パターン数を計算するときに,ある範囲の和が欲しくなるが,しゃくとり法を用いたり,累積和を計算しておいて二分探索で範囲を求めるなどすることができる.
時間計算量 $O(N \log N)$ 程度.

C++によるスパゲッティなソースコード

#include<bits/stdc++.h>
using namespace std;

#define REP(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

#define mygc(c) (c)=getchar_unlocked()
#define mypc(c) putchar_unlocked(c)

#define ll long long
#define ull unsigned ll

void reader(int *x){int k,m=0;*x=0;for(;;){mygc(k);if(k=='-'){m=1;break;}if('0'<=k&&k<='9'){*x=k-'0';break;}}for(;;){mygc(k);if(k<'0'||k>'9')break;*x=(*x)*10+k-'0';}if(m)(*x)=-(*x);}
void writer(int x, char c){int s=0,m=0;char f[10];if(x<0)m=1,x=-x;while(x)f[s++]=x%10,x/=10;if(!s)f[s++]=0;if(m)mypc('-');while(s--)mypc(f[s]+'0');mypc(c);}
template<class T> void writerLn(T x){writer(x,'\n');}

char memarr[17000000]; void *mem = memarr;
#define MD 1000000007

int get_inv(ll a, int md){ll t=a,s=md,u=1,v=0,e;while(s){e=t/s;t-=e*s;u-=e*v;swap(t,s);swap(u,v);}if(u<0)u+=md;return u;}

struct mint{
  static unsigned md, W, R, Rinv, mdninv, RR;
  unsigned val;

  mint(){}mint(int a){val=mulR(a);}mint(unsigned a){val=mulR(a);}mint(ll a){val=mulR(a);}mint(ull a){val=mulR(a);}
  unsigned setmod(unsigned m){int i;unsigned t;W=32;md=m;R=(1ULL<<W)%md;RR=(ull)R*R%md;switch(m){case 104857601:Rinv=2560000;mdninv=104857599;break;case 998244353:Rinv=232013824;mdninv=998244351;break;case 1000000007:Rinv=518424770;mdninv=2226617417U;break;case 1000000009:Rinv=171601999;mdninv=737024967;break;case 1004535809:Rinv=234947584;mdninv=1004535807;break;case 1007681537:Rinv=236421376;mdninv=1007681535;break;case 1012924417:Rinv=238887936;mdninv=1012924415;break;case 1045430273:Rinv=254466304;mdninv=1045430271;break;case 1051721729:Rinv=257538304;mdninv=1051721727;break;default:Rinv=get_inv(R,md);mdninv=0;t=0;rep(i,W){if(t%2==0)t+=md,mdninv|=(1U<<i);t/=2;}}}
  unsigned mulR(unsigned a){return(ull)a*R%md;}unsigned mulR(int a){if(a<0)a=a%md+md;return mulR((unsigned)a);}unsigned mulR(ull a){return mulR((unsigned)(a%md));}unsigned mulR(ll a){if(a<0)a=a%md+md;return mulR((unsigned)a);}
  unsigned reduce(unsigned T){unsigned m=T*mdninv;unsigned t=(unsigned)((T+(ull)m*md)>>W);if(t>=md)t-=md;return t;}unsigned reduce(ull T){unsigned m=(unsigned)T*mdninv;unsigned t=(unsigned)((T+(ull)m*md)>>W);if(t>=md)t-=md;return t;}
  unsigned get(){return reduce(val);}
  mint&operator+=(mint a){val+=a.val;if(val>=md)val-=md;return*this;}mint&operator-=(mint a){if(val<a.val)val=val+md-a.val;else val-=a.val;return*this;}mint&operator*=(mint a){val=reduce((ull)val*a.val);return*this;}mint&operator/=(mint a){return*this*=a.inverse();}
  mint operator+(mint a){return mint(*this)+=a;}mint operator-(mint a){return mint(*this)-=a;}mint operator*(mint a){return mint(*this)*=a;}mint operator/(mint a){return mint(*this)/=a;}
  mint operator+(int a){return mint(*this)+=mint(a);}mint operator-(int a){return mint(*this)-=mint(a);}mint operator*(int a){return mint(*this)*=mint(a);}mint operator/(int a){return mint(*this)/=mint(a);}
  mint operator+(ll a){return mint(*this)+=mint(a);}mint operator-(ll a){return mint(*this)-=mint(a);}mint operator*(ll a){return mint(*this)*=mint(a);}mint operator/(ll a){return mint(*this)/=mint(a);}
  mint operator-(void){mint res;if(val)res.val=md-val;else res.val=0;return res;}
  operator bool(void){return val!=0;}operator int(void){return get();}operator ll(void){return get();}
  mint inverse(){int a=val,b=md,u=1,v=0,t;mint r;while(b){t=a/b;a-=t*b;swap(a,b);u-=t*v;swap(u,v);}if(u<0)u+=md;r.val=(ull)u*RR%md;return r;}
  mint pw(ull b){mint a(*this),r;r.val=R;while(b){if(b&1)r*=a;b>>=1;a*=a;}return r;}
  bool operator==(int a){return get()==a;}
  bool operator!=(int a){return get()!=a;}
};
unsigned mint::md, mint::W, mint::R, mint::Rinv, mint::mdninv, mint::RR;
mint operator+(int a, mint b){return mint(a)+=b;}mint operator-(int a, mint b){return mint(a)-=b;}mint operator*(int a, mint b){return mint(a)*=b;}mint operator/(int a, mint b){return mint(a)/=b;}
mint operator+(ll a, mint b){return mint(a)+=b;}mint operator-(ll a, mint b){return mint(a)-=b;}mint operator*(ll a, mint b){return mint(a)*=b;}mint operator/(ll a, mint b){return mint(a)/=b;}

mint mval[10000], minv[10000];
void mint_init(int md=MD, mint val[]=mval, int vals=1000, mint inv[]=minv, int invs=1000){int i;val[0].setmod(md);val[0].val=0;REP(i,1,vals){val[i].val=val[i-1].val+mint::R;if(val[i].val >=md)val[i].val-=md;}inv[1].val=1;REP(i,2,invs){inv[i].val=md-((ll)(md/i)*inv[md%i].val%md);}REP(i,1,invs)inv[i].val=(ull)inv[i].val*mint::R%md;}

template<class T> void *malloc1d(T **arr, int x, void *mem){(*arr)=(T*)mem;return((*arr)+x);}
template<class T> void *malloc2d(T ***arr, int x, int y, void *mem){int i;(*arr)=(T**)mem;(*arr)[0]=(T*)((*arr)+x);REP(i,1,x)(*arr)[i]=(*arr)[i-1]+y;return((*arr)[x-1]+y);}

int N;
int D[100000], cnt[100001];

mint *dp, *nx;

int main(){
  int i, j, k;
  mint tmp;

  mint_init();

  reader(&N);
  rep(i,N) reader(&k), cnt[k]++;
  k = 0;
  rep(i,100001) while(cnt[i]--) D[k++] = i;

  mem = malloc1d(&dp, N, mem);
  mem = malloc1d(&nx, N, mem);

  rep(i,N) dp[i] = mval[1];
  rep(k,3){
    tmp = mval[0];
    j = 0;
    rep(i,N){
      while(2*D[j] <= D[i]) tmp += dp[j++];
      nx[i] = tmp;
    }
    swap(dp, nx);
  }

  tmp = mval[0];
  rep(i,N) tmp += dp[i];
  writerLn((int)tmp);

  return 0;
}

Current time: 2017年09月21日04時53分05秒
Last modified: 2015年08月16日04時35分35秒 (by laycrs)
Tags: Competitive_Programming AtCoder AtCoder_Regular_Contest ARC043 ARC_B
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