HDU 6333 分块 | 莫队算法

题目链接

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题意：
给定$n,m$，求出$\sum_{i = 0}^m C_{n}^i$，一共有$1e5$组询问。

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思路：

记

$$F[n][m] = \sum_{i = 0}^m C_{n}^i$$

则结合杨辉三角，可得：

$$F[n+1][m] = 2 * F[n][m] - C_n ^ m$$
$$F[n-1][m] = (F[n][m] + C_{n-1}^m) / 2$$
$$F[n][m+1] = F[n][m] + C_n ^ {m+1}$$
$$F[n][m-1] = F[n][m] - C_n ^ m$$

故可以直接使用莫队算法来解决该题，时间复杂度 $O(n \sqrt n)$

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也可以直接分块，因为$n,m \leq 1e5$，故所有的$F$共有$1e10$个数，因为每一个维度的$+1/-1$均可以$O(1)$推出，所以可以直接对$n$分块，假定块的大小为$nuit = 500$，则时间复杂度为：$O(n*m/nuit)$

此题得解。

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代码：
莫队算法：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;

const int mod = 1e9 + 7;
const int A = 1e5 + 10;
int fac[A], Inv[A], Ans[A], block, T;
ll res;

class Que{
public:
    int L, R, id;

    bool operator<(const Que& rhs) const{
        if(L/block == rhs.L/block) return R < rhs.R;
        return L/block < rhs.L/block;
    }
}Q[A];

int Mul(ll a, int b){
    a *= b;
    return a >= mod?a%mod:a;
}

int Add(int a, int b){
    a += b;
    return a >= mod?a-mod:a;
}

int Sub(int a, int b){
    a -= b;
    return a < 0?a+mod:a;
}

int fast_mod(int n, int m){
    int res = 1;
    while(m > 0){
        if(m&1) res = Mul(res, n);
        n = Mul(n, n);
        m >>= 1;
    }
    return res;
}

int Comb(int n, int m){
    return Mul(Mul(fac[n], Inv[m]), Inv[n-m]);
}

void Init(){
    fac[0] = Inv[0] = 1;
    for (int i = 1; i < A; i++) fac[i] = Mul(fac[i-1],i);
    Inv[A-1] = fast_mod(fac[A-1], mod - 2);
    for (int i = A-2; i>=1; i--) Inv[i] = Mul(Inv[i+1], i + 1);
}

void solve(){
    int l = 1, r = 0;
    res = 1;
    for (int i = 1; i <= T; i++) {
        while (r < Q[i].R) {
            r++;
            res = Sub(Mul(res, 2), Comb(r - 1, l));
        }
        while (l > Q[i].L) {
            res = Sub(res, Comb(r, l));
            l--;
        }
        while (r > Q[i].R) {
            res = Mul(Add(res, Comb(r - 1, l)), Inv[2]);
            r--;
        }
        while (l < Q[i].L) {
            l++;
            res = Add(res, Comb(r, l));
        }
        Ans[Q[i].id] = res;
    }
    for (int i = 1; i <= T; i++) printf("%d\n", Ans[i]);
}

int main(){
    Init();
    scanf("%d",&T);
    block = 0;
    for (int i = 1; i <= T; i++) {
        Q[i].id = i;
        scanf("%d%d", &Q[i].R, &Q[i].L);
        block = max(block, Q[i].R);
    }
    block = (int)sqrt(1.0 * block);
    sort(Q + 1, Q + 1 + T);

    solve();
    return 0;
}

对分块：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
typedef long long ll;

const int mod = 1e9 + 7;
const int A = 1e5 + 1000;
const int B = A/500 + 100;
int fac[A], Inv[A];
int a[B][A], tem[2][A];

int Mul(ll a, int b){
    a *= b;
    return a >= mod?a%mod:a;
}

int Add(int a, int b){
    a += b;
    return a >= mod?a-mod:a;
}

int Sub(int a, int b){
    a -= b;
    return a < 0?a+mod:a;
}

int fast_mod(int n, int m){
    int res = 1;
    while(m > 0){
        if(m&1) res = Mul(res, n);
        n = Mul(n, n);
        m >>= 1;
    }
    return res;
}

int Comb(int n, int m){
    return Mul(Mul(fac[n], Inv[m]), Inv[n-m]);
}

void Init(){
    fac[0] = Inv[0] = 1;
    for (int i = 1; i < A; i++) fac[i] = Mul(fac[i-1],i);
    Inv[A-1] = fast_mod(fac[A-1], mod - 2);
    for (int i = A-2; i>=1; i--) Inv[i] = Mul(Inv[i+1], i + 1);

    for (int i = 0; i < A; i += 500) {
        a[i/500][0] = 1;
        for (int j = 1; j <= i; j++) a[i/500][j] = Add(a[i/500][j-1], Comb(i,j));
    }
}

int main(){
    Init();
    int T, n, m, res;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d", &n, &m);
        n++;
        res = 0;
        res = a[(n/500)+1][m];
        for (int i = ((n/500)+1)*500; i >= n; i--){
            res = Mul(Add(res,Comb(i-1,m)), Inv[2]);
        }
        printf("%d\n", res);
    }
    return 0;
}