Problem B. Harvest of Apples 莫队求组合数前缀和

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≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).

 

 

Output

For each test case, print an integer representing the number of ways modulo 109+7.

 

 

Sample Input

2 5 2 1000 500

 

 

Sample Output

16 924129523

 

 

Source

2018 Multi-University Training Contest 4

 

 

Recommend

 

 

这题刚写的时候 公式及其的容易推 C(n,0)+C(n,1)+C(n,2)+。。。+C(n,m)

然后一直在想能不能化简这一项一项的求 复杂度会爆炸的

 

最后的题解是莫队

C(n,m)=C(n-1,m-1)+C(n-1,m)   

设 S(n,m)=C(n,0)+C(n,1)+C(n,2)+。。。+C(n,m)

然后将S(n,m) 通过 第一个公式 拆项

最后化简 变为 S(n,m)=2*S(n-1,m)-C(n-1,m);

 

预处理阶乘逆元  然后就 OK了 

莫队大法好

 

 

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-6
#define  fi first
#define  se second
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
#define  bug         printf("******\n")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  fuck(x)     cout<<"["<<x<<"]"<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("DATA.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#pragma  comment (linker,"/STACK:102400000,102400000")
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int t, sz;
LL inv[maxn], a[maxn], b[maxn];
struct node {
    int l, r, id;
    LL ans = 0;
} qu[maxn];
int cmp(node a, node b) {
    return a.l / sz == b.l / sz ? a.r < b.r : a.l < b.l;
}
LL expmod(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b = b >> 1;
    }
    return ans;
}
void init() {
    a[1] = 1;
    for (int i = 2 ; i < maxn ; i++) a[i] = a[i - 1] * i % mod;
    for (int i = 1 ; i < maxn ; i++) b[i] = expmod(a[i], mod - 2);
}
LL C(int n, int m) {
    if (m > n || n < 0 || m < 0 ) return 0;
    if (m == n || m == 0) return 1;
    return a[n] * b[m] % mod * b[n - m] % mod;
}

int main() {
    init();
    sf(t);
    for (int i = 1 ; i <= t ; i++) {
        sff(qu[i].l, qu[i].r);
        qu[i].id = i, qu[i].ans = 0;
    }
    sz = sqrt(maxn);
    sort(qu + 1, qu + 1 + t, cmp);
    LL sum = 1;
    for (int i = 1, L = 1, R = 0 ; i <= t ; i++) {
        while(L < qu[i].l) sum = (2 * sum - C(L++, R) + mod) % mod;
        while(L > qu[i].l) sum = ((sum + C(--L, R)) * b[2]) % mod;
        while(R < qu[i].r) sum = (sum + C(L, ++R)) % mod;
        while(R > qu[i].r) sum = (sum - C(L, R--) + mod) % mod;
        qu[qu[i].id].ans = sum;
    }
    for (int i = 1 ; i <= t ; i++) printf("%lld\n", qu[i].ans);
    return 0;
}