2018 Multi-University Training Contest 4

莫队 + 组合数

令f(n, m)表示从n个苹果中拿不超过m个的方法数。

则 f(n, m) = f(n, m - 1) + C(n, m)

则 f(n + 1, m) = f(n, m) + f(n, m - 1) = 2f(n, m) - C(n, m) (相当于第n+1个苹果不取的方法数+取的方法数)

推导到此，可以发现f(n, m + 1), f(n, m - 1), f(n + 1, m), f(n - 1, m)可由f(n, m)用O(1)的时间推出。

那么我们可以把(m, n)看成时间轴的两端，用莫队移动即可

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 100005;
const int MOD = 1e9 + 7;
int _, t;
ll ans, res[N], fac[N], inv[N];
struct Query{
    int n, m, id, block;
    bool operator < (const Query &rhs) const {
        return (block ^ rhs.block) ? m < rhs.m : (block ^ 1) ? n < rhs.n : n > rhs.n;
    }
}query[N];

void calc(){
    fac[0] = 1, fac[1] = 1, inv[0] = 1, inv[1] = 1;
    for(int i = 2; i <= N; i ++){
        fac[i] = (fac[i - 1] * i) % MOD;
        inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
    }
    for(int i = 1; i <= N; i ++){
        inv[i] = (inv[i] * inv[i - 1]) % MOD;
    }
    t = (int)sqrt(N);
}

ll C(int n, int m){
    return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}

int main(){

    calc();
    _ = read();
    for(int i = 1; i <= _; i ++){
        query[i].n = read(), query[i].m = read();
        query[i].id = i, query[i].block = (query[i].m - 1) / t + 1;
    }
    sort(query + 1, query + _ + 1);
    int l = 1, r = 0;
    ans = 1;
    for(int i = 1; i <= _; i ++){
        int curL = query[i].m, curR = query[i].n;
        while(r < curR){
            r ++;
            ans = ((2 * ans - C(r - 1, l)) % MOD + MOD) % MOD;
        }
        while(l > curL){
            l --;
            ans = ((ans - C(r, l + 1)) % MOD + MOD) % MOD;
        }
        while(r > curR){
            ans = ((ans + C(r - 1, l)) % MOD * inv[2]) % MOD;
            r --;
        }
        while(l < curL){
            ans = (ans + C(r, l + 1)) % MOD;
            l ++;
        }
        res[query[i].id] = ans;
    }
    for(int i = 1; i <= _; i ++){
        printf("%lld
", res[i]);
    }
    return 0;
}