WA on #2 求调,能过样例QAQ
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WA on #2 求调,能过样例QAQ
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Stevehim楼主2023/9/1 17:25
#include <bits/stdc++.h>
#define maxn 400010
#define reg register
#define ll long long
using namespace std;
ll fac[maxn]; //计算阶乘
ll inv[maxn];
const int mod = 1000000007;
namespace IO {
	char *p1, *p2, buf[50];
//#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,50,stdin),p1==p2)?EOF:*p1++)
#define nc() getchar()
	inline int read() {
		int x = 0, f = 1;
		char ch = nc();
		while (ch < 48 || ch > 57) {
			if (ch == '-')
				f = -1;
			ch = nc();
		}
		while (ch >= 48 && ch <= 57)
			x = (x << 1) + (x << 3) + (ch ^ 48),
			ch = nc();
		return x * f;
	}
}
using IO::read;
ll quickpow(ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) {
			ans = ans * a % mod;
		}
		a = a * a % mod;
		b >>= 1;
	}
	return ans % mod;
}
int n;
void init() {
	fac[0] = inv[0] = 1;
	for (reg int i = 1; i < maxn; i++) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = inv[i - 1] * quickpow(i, mod - 2) % mod;
	}
}
ll C(int a, int b) {
	return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
struct node {
	int qr, ql, qi;
} q[maxn];
int bl;
inline bool cmp(node a,node b){  //比较函数,以左端点所在块为第一关键字
	return a.ql / bl == b.ql / bl ? (a.ql / bl % 2 ? a.qr > b.qr : a.qr < b.qr) : a.ql < b.ql;  
}
ll ans;
ll an[maxn];
int nown, nowk;

inline void addn() {
	ans = ((ans * 2 % mod) - C(nown, nowk));
	++nown;
}

inline void subn() {
	ans = (ans + C(nown - 1, nowk)) % mod * inv[2] % mod;
	--nown;
}

inline void addk() {
	ans = (ans + C(nown, nowk + 1)) % mod;
	++nowk;
}

inline void subk() {
	ans = ((ans - C(nown, nowk))% mod + mod) % mod;
	--nowk;
}
int m, k;
int main() {
//	freopen("1.in","r",stdin);
	init();
	scanf("%d",&m);
	bl = 317;
	for (reg int i = 1; i <= m; i++)
		scanf("%d %d",&q[i].ql,&q[i].qr), q[i].qi = i;
	sort(q + 1, q + 1 + m, cmp);
	nown = 1, nowk = 0, ans = 1;
	for (reg int i = 1; i <= m; i++) {
		while (nown < q[i].ql) addn();
		while (nown > q[i].ql) subn();
		while (nowk < q[i].qr) addk();
		while (nowk > q[i].qr) subk();
//		printf("%lld\n",ans);
		an[q[i].qi] = ans;
	}
	for (reg int i = 1; i <= m; i++) {
		printf("%lld\n", an[i]);
	}
	return 0;
}
2023/9/1 17:25
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