2018hdu杭电多校第四场 hdu6333 Problem B. Harvest of Apples (莫队 + 组合数 + 逆元)

Problem B. Harvest of Apples

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1595    Accepted Submission(s): 603

Problem Description

There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.

Input

The first line of the input contains an integer T (1≤T≤) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤).

Output

For each test case, print an integer representing the number of ways modulo .

Sample Input

2

5 2

1000 500

Sample Output

16

924129523

Source

2018 Multi-University Training Contest 4

---

题意：t组数据，每组给出n和m，求（）

题解：

组合数求模要用到逆元， 此时称 c 为 b 的逆元

假设c是b的逆元  ①,根据费马小定理： 时， ② （p表示素数）

联立①②可得：  所以模（p-2）就好

然后回到这题，显而易见可以得到

根据组合数的递推公式，

把上式的变成或者 得到:

根据红字的公式用莫队离线操作

代码：

#include 
#define mem(a, x) memset(a, x, sizeof(a))
#define rep(i,a,n) for (int i=a;i=a;i--)
using namespace std;
typedef long long ll;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
//------------------------------------head------------------------------------
const int N = 1e5+10;
ll fac[N], inv[N]; // fac 阶乘， inv 逆元 
ll inv2, val, ans[N]; // inv2 除2的逆元，val 中间答案，ans 答案 
int pos[N], t; // pos 分块 

void init() {
    // 除2的逆元 
    inv2 = powmod(2, mod-2); 
    
    // 预处理阶乘 
    fac[0] = fac[1] = 1;
    rep (i, 2, N) fac[i] = i * fac[i-1] % mod;
    
    // 求阶乘逆元 
    inv[N-1] = powmod(fac[N-1], mod-2);
	per (i, 0, N-1) inv[i] = inv[i+1] * (i+1) % mod;
}

ll comb(int n, int m) { // c(n, m) = n!/(m! * (n-m)!)
	return fac[n] * inv[m] % mod * inv[n-m] % mod;
}

struct node {
	int l, r, id;
} Q[N];

bool cmp(node a, node b) {
	if (pos[a.l] == pos[b.l]) return a.r < b.r;
	return a.l < b.l;
}

void addL(int n, int m) { // S(n,m) = 2 * S(n-1,m) - C(n-1,m)
	val = (2 * val % mod - comb(n-1, m) + mod) % mod;
}

void delL(int n, int m) { // S(n-1,m) = (S(n,m) + C(n-1,m)) / 2 
	val = (val + comb(n-1, m)) % mod * inv2 % mod;
}

void addR(int n, int m) { // S(n,m) = S(n,m-1) + C(n,m)
	val = (val + comb(n, m)) % mod;
}

void delR(int n, int m) { // S(n,m-1) = S(n,m) - C(n, m)
	val = (val - comb(n,m) + mod) % mod;
}

int main() {
	init();
	scanf("%d", &t);
	int sz = sqrt(N);
	rep (i, 1, t+1) {
		scanf("%d%d", &Q[i].l, &Q[i].r);
		pos[i] = i / sz;
		Q[i].id = i;
	}
	sort(Q+1, Q+t+1, cmp);
	int l = 1, r = 1;
	val = 2; // c(1,0) + c(1,1)
	rep (i, 1, t+1) {
		while (l < Q[i].l) addL(++l, r);
		while (l > Q[i].l) delL(l--, r);
		while (r < Q[i].r) addR(l, ++r);
		while (r > Q[i].r) delR(l, r--);
		ans[Q[i].id] = val;
	}
	rep (i, 1, t+1) printf("%lld\n", ans[i]);
}