题目:Problem B. Harvest of Apples
题意:给定n,m,求 ;
思路:设S(n,m) = ,那么S(n,k)=S(n,k-1)+C(n,k); S(n,k)=2*S(n-1,k)-C(n-1,k);由此,我们得到了S(n,k),S(n-1,k),S(n,k-1)这3个状态互相转移的方程,使用莫队求解。莫队复杂度为O(nsqrt(n)),本题中m<=n,所以用m来分块更优。
#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAX = 1e5+10;
const int MOD=1e9+7;
typedef struct{
int n,m;
int id;
int block;//所属块
}Point;
Point a[MAX];
ll out[MAX];
//****组合数模板****
ll F[MAX], Finv[MAX], inv[MAX];//F是阶乘,Finv是逆元的阶乘
void init(){
inv[1] = 1;
for(int i = 2; i < MAX; i ++){
inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;
}
F[0] = Finv[0] = 1;
for(int i = 1; i < MAX; i ++){
F[i] = F[i-1] * 1ll * i % MOD;
Finv[i] = Finv[i-1] * 1ll * inv[i] % MOD;
}
}
ll comb(int n, int m){//comb(n, m)就是C(n, m)
if(m < 0 || m > n) return 0;
return F[n] * 1ll * Finv[n - m] % MOD * Finv[m] % MOD;
}
//******************
bool cmp(Point x,Point y)
{
return x.block==y.block?x.n<y.n:x.m<y.m;
}
int main()
{
int T;
scanf("%d",&T);
init();
int unit=sqrt(1e5); //块大小
for(int i=1;i<=T;i++)
{
scanf("%d%d",&a[i].n,&a[i].m);
a[i].id=i;
a[i].block=a[i].m/unit; //根据m分块
}
sort(a+1,a+T+1,cmp);
int nn=1,mm=0;
ll ans=1;
for(int i=1;i<=T;i++)
{
while(nn<a[i].n){
ans=(2*ans%MOD-comb(nn,mm)+MOD)%MOD;
nn++;
}
while(nn>a[i].n){
nn--;
ans=(ans+comb(nn,mm))%MOD*inv[2]%MOD;
}
while(mm<a[i].m){
mm++;
ans=(ans+comb(nn,mm))%MOD;
}
while(mm>a[i].m){
ans=(ans-comb(nn,mm)+MOD)%MOD;
mm--;
}
out[a[i].id]=ans;
}
for(int i=1;i<=T;i++)
printf("%lld\n",out[i]);
return 0;
}