hdu6333 /// 莫队+组合数

题目大意：

给定n m

在n个数中最多选择m个的所有方案

题解：https://blog.csdn.net/qq_36300700/article/details/81385244

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
const int mod=1e9+7;
const int N=1e5+5;

/********组合数模板*********/
LL pow_mod(LL a, LL b) {
    LL res = 1LL; a %= mod;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    } return res;
}
LL inv(LL a) { return pow_mod(a, mod-2); }
LL F[N], Finv[N];//F是阶乘，Finv是逆元的阶乘
void init() {
    F[0] = Finv[0] = 1LL;
    for(int i = 1; i < N; i ++){
        F[i] = F[i-1] * 1LL * (LL)i % mod;
        Finv[i] = Finv[i-1] * 1LL * inv(i) % mod;
    }
} // O(n)预处理
LL C(LL n, LL m) {
    if(m < 0 || m > n) return 0;
    return F[n] * 1LL * Finv[n - m] % mod * Finv[m] % mod;
} // O(1)获得组合数C(n,m)
/**************************/

LL res[N];

/********莫队*********/
int len;
struct Q {
    LL n,m;
    int block, id;
    bool operator <(const Q& q)const {
        if(block==q.block) return n<q.n;
        return block<q.block;
    }
}q[N];
void Mo() { 
    LL L=0, R=0, ans=1LL;
    for(int i=0;i<t;i++) {
        LL l=q[i].n, r=q[i].m;
        while(L>l) ans=((ans+C(L-1LL,R))%mod*Finv[2])%mod, L--;
        while(L<l) ans=(2*ans%mod-C(L,R)+mod)%mod, L++;
        while(R<r) ans=(ans+C(L,R+1))%mod, R++;
        while(R>r) ans=(ans-C(L,R)+mod)%mod, R--;
        res[q[i].id]=ans;
    }
}
/**************************/

int main()
{
    init(); len=sqrt(N);
    int t; scanf("%d",&t);
    for(int i=0;i<t;i++) {
        scanf("%lld%lld",&q[i].n,&q[i].m);
        q[i].block=q[i].m/len; q[i].id=i;
    }
    sort(q,q+t);
    Mo();
    for(int i=0;i<t;i++)
        printf("%lld
",res[i]);

    return 0;
}

莫队模板

int len;
struct Q {
    LL l,r;
    int block, id;
    bool operator <(const Q& q)const {
        if(block==q.block) return n<q.n;
        return block<q.block;
    }
}q[N];
void Mo() { 
    len=sqrt(N);
    sort(q,q+t);
    LL L=0, R=0, ans= ;
    for(int i=0;i<t;i++) {
        LL l=q[i].l, r=q[i].r;
        while(L>l) --L, ans= ;
        while(L<l) L++, ans= ; 
        while(R>r) R--, ans= ;
        while(R<r) ++R, ans= ;
        res[q[i].id]=ans;
    }
}