【2018多校第四场 hdu 6333】【莫队分块】【思路好题】【思维】【求C(n,0)+C(n,1)+……+C(n,m)的值】

【链接】

http://acm.hdu.edu.cn/showproblem.php?pid=6333

【题意】

求C(n,0)+C(n,1)+……+C(n,m)的值。

【思路】k

由于t,n都比较大，需要查询的复杂度不能太高，需要离线处理，由组合数的将前k项求和可以推出如果已知一个S(n,m),可以轻松得出S(n-1,m),S(n+1,m),S(n,m-1),S(n,m+1)。容易想到莫队分块算法。将n,m转换成莫队的区间查询，将n当成l，m当成r即可

【代码】

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const int MOD = 1e9 + 7;
const int maxn = 1e5 + 6;
int unit ;

struct node {
	int l, r, id;
	inline node(){}
	inline node(int l,int r,int id):l(l),r(r),id(id){}
	inline bool operator <(const node &a) const {
		if (l / unit == a.l / unit)return r < a.r;
		else return l / unit < a.l / unit;
	}
}k[maxn];

ll ans[maxn];
ll fac[maxn];
ll invfac[maxn];
ll inv[maxn];

inline void init() {
	fac[0] = invfac[0] = 1;
	fac[1] = inv[1] = invfac[1] = 1;
	for (int i = 2; i < maxn; ++i) {
		fac[i] = fac[i - 1] * i%MOD;
		inv[i] = inv[MOD%i] * (MOD - MOD / i) % MOD;
		invfac[i] = invfac[i - 1] * inv[i] % MOD;
	}
}

inline ll cal(int a, int b) {
	ll res = fac[a] * invfac[b] % MOD*invfac[a - b] % MOD;
	return res;
}
int n;

inline void work() {
	ll tmp = 0;
	for (int i = 0; i <= k[1].r; ++i) {
		tmp = (tmp + cal(k[1].l, i)) % MOD;
	}
	ans[k[1].id] = tmp;
	int L = k[1].l;
	int R = k[1].r;
	for (int i = 2; i <= n; i++) {
		while (L < k[i].l) {
			tmp = (tmp * 2%MOD - cal(L++, R) + MOD) % MOD;
		}
		while (L > k[i].l) {
			tmp = (tmp + cal(--L, R) + MOD) % MOD*inv[2] % MOD;
		}
		while (R < k[i].r) {
			tmp = (tmp + cal(L, ++R)) % MOD;
		}
		while (R > k[i].r) {
			tmp = (tmp - cal(L, R--) + MOD) % MOD;
		}
		ans[k[i].id] = tmp;
	}
}

int main() {
	init();
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &k[i].l, &k[i].r);
		k[i].id = i;
	}
	unit =(int) sqrt(n); 
	sort(k + 1, k + 1 + n);
	work();
	for (int i = 1; i <= n; i++) {
		printf("%d\n", ans[i]);
	}
}