結果

問題 No.1496 不思議な数え上げ
ユーザー akuaakua
提出日時 2023-03-04 20:08:54
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 665 ms / 3,000 ms
コード長 6,511 bytes
コンパイル時間 20,189 ms
コンパイル使用メモリ 251,732 KB
最終ジャッジ日時 2025-02-11 05:24:12
ジャッジサーバーID
(参考情報)
judge1 / judge6
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 39
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <atcoder/all>
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <iomanip>
#include <functional>
using namespace std;
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repi(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
typedef long long ll;
typedef unsigned long long ull;
const ll inf=1e18;
using graph = vector<vector<int> > ;
using P= pair<ll,ll>;
using vi=vector<int>;
using vvi=vector<vi>;
using vll=vector<ll>;
using vvll=vector<vll>;
using vp=vector<P>;
using vpp=vector<vp>;
using vd=vector<double>;
using vvd =vector<vd>;
//string T="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
//string S="abcdefghijklmnopqrstuvwxyz";
//g++ main.cpp -std=c++17 -I .
//cout <<setprecision(20);
//cout << fixed << setprecision(10);
//cin.tie(0); ios::sync_with_stdio(false);
const double PI = acos(-1);
int vx[]={0,1,0,-1,-1,1,1,-1},vy[]={1,0,-1,0,1,1,-1,-1};
void putsYes(bool f){cout << (f?"Yes":"No") << endl;}
void putsYES(bool f){cout << (f?"YES":"NO") << endl;}
void putsFirst(bool f){cout << (f?"First":"Second") << endl;}
void debug(int test){cout << "TEST" << " " << test << endl;}
ll pow_pow(ll x,ll n,ll mod){
if(n==0) return 1;
x%=mod;
ll res=pow_pow(x*x%mod,n/2,mod);
if(n&1)res=res*x%mod;
return res;
}
ll gcd(ll x,ll y){
if(y==0)return x;
return gcd(y,x%y);
}
ll lcm(ll x,ll y){
return ll(x/gcd(x,y))*y;
}
template<class T> bool chmin(T& a, T b) {
if (a > b) {
a = b;
return true;
}
else return false;
}
template<class T> bool chmax(T& a, T b) {
if (a < b) {
a = b;
return true;
}
else return false;
}
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
ll mod =998244353;
struct mint {
ll x; // typedef long long ll;
mint(ll x=0):x((x%mod+mod)%mod){}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
mint pow(ll t) const {
if (!t) return 1;
mint a = pow(t>>1);
a *= a;
if (t&1) a *= *this;
return a;
}
// for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
vector<mint> fact, ifact;
combination(int n):fact(n+1),ifact(n+1) {
//assert(n < mod);
fact[0] = 1;
for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
}
mint operator()(int n, int k) {
if (k < 0 || k > n) return 0;
return fact[n]*ifact[k]*ifact[n-k];
} mint p(int n, int k) { return fact[n]*ifact[n-k]; } } c(500000);
using vm=vector<mint> ;
using vvm=vector<vm> ;
ll sqrt_(ll x) {
ll l = 0, r = ll(3e9)+1;
while (l+1<r) {
ll c = (l+r)/2;
if (c*c <= x) l = c; else r = c;
}
return l;
}
int valid(int x,int y,int h,int w){
if(x>=0 && y>=0 && x<h && y<w)return 1;
else return 0;
}
struct edge{
int to; ll cost;
edge(int to,ll cost) : to(to),cost(cost){}
};
//snuke
template<typename T>
struct Matrix {
int h, w;
vector<vector<T> > d;
Matrix() {}
Matrix(int h, int w, T val=0): h(h), w(w), d(h, vector<T>(w,val)) {}
Matrix& unit() {
//assert(h == w);
rep(i,h) d[i][i] = 1;
return *this;
}
const vector<T>& operator[](int i) const { return d[i];}
vector<T>& operator[](int i) { return d[i];}
Matrix operator*(const Matrix& a) const {
//assert(w == a.h);
Matrix r(h, a.w);
rep(i,h)rep(k,w)rep(j,a.w) {
r[i][j] += d[i][k]*a[k][j];
}
return r;
}
Matrix pow(ll t) const {
// assert(h == w);
if (!t) return Matrix(h,h).unit();
if (t == 1) return *this;
Matrix r = pow(t>>1);
r = r*r;
if (t&1) r = r*(*this);
return r;
}
};
using ve=vector<edge>;
using vve=vector<ve>;
//g++ main.cpp -std=c++17 -I .
int main(){cin.tie(0);ios::sync_with_stdio(false);
int n; cin >> n;
vi p(n+1);vll a(n+1);
vi id(n+1);
repi(i,1,n+1)cin >> p[i];
repi(i,1,n+1)id[p[i]]=i;
repi(i,1,n+1)cin >> a[i];
vll sum(n+1);
repi(i,1,n+1)sum[i]=sum[i-1]+p[i];
vi q(n+1);
for(int i=1; i<=n; i++)q[i]=p[n+1-i];
vll sum2(n+1);
for(int i=1; i<=n; i++)sum2[i]=sum2[i-1]+q[i];
set<int> st;
st.insert(0);
st.insert(n+1);
vll anss;
for(int minv=1; minv<=n; minv++){
int ce=id[minv];
auto it=st.lower_bound(ce);
int r=(*it)-1;
it--;
int l=(*it)+1;
ll ans=0;
if(a[minv]>=minv){
if(ce-l<=r-ce){
for(int i=l; i<=ce; i++){
int index=upper_bound(sum.begin(),sum.end(),sum[i-1]+a[minv])-sum.begin();
index--;
ans+=max(0,min(r,index)-ce+1);
}
}
else {
int ce2=n+1-ce;
int l2=n+1-r;
int r2=n+1-l;
for(int i=l2; i<=ce2; i++){
int index=upper_bound(sum2.begin(),sum2.end(),sum2[i-1]+a[minv])-sum2.begin();
index--;
ans+=max(0,min(r2,index)-ce2+1);
}
}
}
st.insert(ce);
anss.push_back(ans);
}
for(auto u:anss)cout << u << endl;
}
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