結果

問題 No.2206 Popcount Sum 2
コンテスト
ユーザー vjudge1
提出日時 2025-08-02 16:15:07
言語 C++14
(gcc 13.3.0 + boost 1.89.0)
結果
TLE  
実行時間 -
コード長 2,934 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 1,836 ms
コンパイル使用メモリ 162,980 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-08-02 16:15:16
合計ジャッジ時間 8,330 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 1 TLE * 1 -- * 16
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int Main()’:
main.cpp:63:19: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘Mint::mint’ [-Wformat=]
   63 |         printf ("%d\n", ans);
      |                  ~^     ~~~
      |                   |     |
      |                   int   Mint::mint

ソースコード

diff #
raw source code 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define il inline
#define N 200005
il int rd(){
	int s = 0, w = 1;
	char ch = getchar();
	for (;ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') w = -1;
	for (;ch >= '0' && ch <= '9'; ch = getchar()) s = ((s << 1) + (s << 3) + ch - '0');
	return s * w;
}
namespace Mint{
	const int P = 998244353;
	class mint {public:
		int num;
		mint() = default;
		mint(int _num) : num((_num % P + P) % P) {}
		mint &operator = (int b) {return *this = mint(b);}
		friend bool operator <(mint a, mint b){return a.num < b.num;}
		friend bool operator >(mint a, mint b){return a.num > b.num;}
		friend bool operator <=(mint a, mint b){return a.num <= b.num;}
		friend bool operator >=(mint a, mint b){return a.num >= b.num;}
		friend bool operator ==(mint a, mint b){return a.num == b.num;}
		friend mint operator +(mint a, mint b){return mint((a.num + b.num) % P);}
		friend mint &operator +=(mint &a, mint b){return a = a + b;}
		friend mint operator +(mint a, int b){return a + mint(b);}
		friend mint &operator +=(mint &a, int b){return a = a + b;}
		friend mint &operator ++(mint &a){return a += 1;}
		friend mint operator ++(mint &a,int){mint copy(a);a += 1;return copy;}
		friend mint operator -(mint a, mint b){return mint(((a.num - b.num) % P + P) % P);}
		friend mint &operator -=(mint &a, mint b){return a = a - b;}
		friend mint operator -(mint a, int b){return a - mint(b);}
		friend mint &operator -=(mint &a, int b){return a = a - b;}
		friend mint &operator --(mint &a){return a -= 1;}
		friend mint operator --(mint &a,int){mint copy(a);a -= 1;return copy;}
		friend mint operator *(mint a, mint b){return mint((ll)a.num * b.num % P);}
		friend mint &operator *=(mint &a, mint b){return a = a * b;}
		friend mint operator *(mint a, int b){return a * mint(b);}
		friend mint &operator *=(mint &a, int b){return a = a * b;}
		mint inv(){ll ans = 1, x = num, r = P - 2;for (; r; x = x * x % P, r >>= 1) if (r & 1) ans = ans * x % P;return ans;}
		friend mint operator /(mint a, mint b){return a * b.inv();}
		friend mint &operator /=(mint &a, mint b){return a = a / b;}
		friend mint operator /(mint a, int b){return a / mint(b);}
		friend mint &operator /=(mint &a, int b){return a = a / b;}
	};
	mint ksm(mint x, ll r){mint ans = 1;for (; r; x = x * x, r >>= 1) if (r & 1) ans = ans * x;return ans;}
}
using namespace Mint;
mint fact[N], inft[N];
il void init(int n){
	fact[0] = 1;
	for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i;
	inft[n] = ksm(fact[n], P - 2);
	for (int i = n; i >= 1; i--) inft[i - 1] = inft[i] * i;
}
il mint C(int n, int m){return (m < 0 || n < m) ? 0 : fact[n] * inft[m] * inft[n - m];}
int n, m;
signed Main(){
	mint ans = 0;
	n = rd(), m = rd();
	for (int i = 0; i <= m; i++) ans += C(n - 1, i - 1) * (ksm(2, n) - 1);
	printf ("%d\n", ans);
	return 0;
}
signed main(){
	init(200005);
	for (int T = rd(); T--;) Main();
	return 0;
}
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