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 + ;
 const int maxn = 1e5 + ;
 int t, sz;
 LL inv[maxn], a[maxn], b[maxn];
 struct node {
     int l, r, id;
     LL ans = ;
 } 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 = ;
     while(b) {
         if (b & ) ans = ans * a % mod;
         a = a * a % mod;
         b = b >> ;
     }
     return ans;
 }
 void init() {
     a[] = ;
     for (int i =  ; i < maxn ; i++) a[i] = a[i - ] * i % mod;
     for (int i =  ; i < maxn ; i++) b[i] = expmod(a[i], mod - );
 }
 LL C(int n, int m) {
     if (m > n || n <  || m <  ) return ;
     if (m == n || m == ) return ;
     return a[n] * b[m] % mod * b[n - m] % mod;
 }

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