結果

問題 No.2336 Do you like typical problems?
ユーザー woodywoodywoodywoody
提出日時 2023-06-02 22:37:54
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 1,087 ms / 2,000 ms
コード長 5,259 bytes
コンパイル時間 4,722 ms
コンパイル使用メモリ 249,784 KB
実行使用メモリ 115,824 KB
最終ジャッジ日時 2024-06-09 00:18:18
合計ジャッジ時間 14,545 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 116 ms
81,532 KB
testcase_01 AC 117 ms
81,368 KB
testcase_02 AC 116 ms
81,304 KB
testcase_03 AC 115 ms
81,536 KB
testcase_04 AC 115 ms
81,536 KB
testcase_05 AC 114 ms
81,408 KB
testcase_06 AC 116 ms
81,536 KB
testcase_07 AC 117 ms
81,296 KB
testcase_08 AC 117 ms
81,664 KB
testcase_09 AC 119 ms
81,704 KB
testcase_10 AC 117 ms
81,752 KB
testcase_11 AC 117 ms
81,792 KB
testcase_12 AC 117 ms
81,912 KB
testcase_13 AC 1,042 ms
115,700 KB
testcase_14 AC 1,087 ms
115,812 KB
testcase_15 AC 1,082 ms
115,824 KB
testcase_16 AC 1,077 ms
115,692 KB
testcase_17 AC 1,087 ms
115,692 KB
testcase_18 AC 245 ms
88,652 KB
testcase_19 AC 255 ms
88,912 KB
testcase_20 AC 656 ms
101,788 KB
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 # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto& i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll

using namespace std;
using namespace atcoder;

using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}}
#define dame { cout << "No" << endl; return;}
#define dame1 { cout << -1 << endl; return;}
#define cout2(x,y) cout << x << " " << y << endl;
#define coutp(p) cout << p.fi << " " << p.se << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define coutv2(v) fore(vy , v) cout << vy << "\n";
#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define coutvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = dynamic_modint<0>;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define yesr { cout <<"Yes"<<endl; return;}
#define no {cout <<"No"<<endl;}
#define nor { cout <<"No"<<endl; return;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#define show(x) cerr<<#x<<" = "<<x<<endl
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show2(x,y) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<endl
#define show3(x,y,z) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<endl
#define show4(x,y,z,x2) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<" : "<<#x2<<" = "<<x2<<endl
#define test(x) cout << "test" << x << endl
#define showv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) fore(vy , v) cout << vy << "\n";
#define showvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) fore(vy , v) cout << vy.val() << "\n";
#else
#define show(x)
#define showp(p)
#define show2(x,y)
#define show3(x,y,z)
#define show4(x,y,z,x2)
#define test(x)
#define showv(v)
#define showv2(v)
#define showvm(v)
#define showvm2(v)
#endif

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
	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];
  }
} com(10000005);

void solve(){
	int n; cin>>n;
	vl bb(n),cc(n);
	rep(i,n) cin>>bb[i]>>cc[i];
	if (n==1){
		cout << 0 << endl;
		return;
	}
	map<ll,int> mp;
	rep(i,n) mp[bb[i]] = 0;
	rep(i,n) mp[cc[i]] = 0;
	int m = 0;
	fore(y , mp) y.se = m++;
	vpll v;
	rep(i,n) v.pb({cc[i],bb[i]});
	sort(rrng(v));
	fenwick_tree<mint> fw1(m),fw2(m);

	mint ans = 0;
	mint sump = 1;
	rep(i,n) sump *= (cc[i]-bb[i]+1);
	rep(i,n){
		int ic = mp[v[i].fi];
		int ib = mp[v[i].se];
		ll b = v[i].se;
		ll c = v[i].fi;
		mint coef = sump/(c-b+1);
		ans += fw1.sum(ib,ic+1)*(c+1)*coef;
		ans -= fw2.sum(ib,ic+1)*coef;
		ans += fw1.sum(0,ib)*(c-b+1)*coef;
		fw1.add(ib,mint(1)/(c-b+1));
		fw2.add(ib,mint(b)/(c-b+1));
	}
	// outm;
	ans *= -1;
	ans += sump*n*(n-1)/2;
	// mint sum = 0;
	mint tmp = 0;
	// rep(i,n){
	// 	mint cb = cc[i]-bb[i]+1;
	// 	sum += cb;
	// 	tmp -= cb*cb;
	// }
	// tmp += (sum*sum);
	// tmp /= 2;
	ans += tmp;
	// outm;
	ans *= com.fact[n-2]*com(n,2);
	tmp = 1;
	rep(i,n) tmp *= (cc[i]-bb[i]+1);
	ans /= tmp;
	outm;


	return;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	int t = 1;
	//cin>>t;

	rep(i,t){
		solve();
	}

	return 0;
}
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