【BZOJ3529】【莫比乌斯反演 + 树状数组】[Sdoi2014]数表

Description

    有一张N×m的数表，其第i行第j列（1 < =i < =礼，1 < =j < =m）的数值为
能同时整除i和j的所有自然数之和。给定a，计算数表中不大于a的数之和。

Input

    输入包含多组数据。
    输入的第一行一个整数Q表示测试点内的数据组数，接下来Q行，每行三个整数n，m，a(|a| < =10^9)描述一组数据。

Output

    对每组数据，输出一行一个整数，表示答案模2^31的值。

Sample Input

2
4 4 3
10 10 5

Sample Output

20
148

HINT

1 < =N．m < =10^5  ， 1 < =Q < =2×10^4

Source

Round 1 Day 1

【分析】

其实是个比较水的题目。

 $sumlimits_{i <  = n { m{ j <  = m}}}^{} { F(gcd(i,j))} $其中gcd(i,j) <= a，F(i)，为i的因子和。

我们可以将公式变形,先忽略gcd(i, j) <= a这个条件。

$sumlimits_{i = 1}^{min (n,m)} {F(i) imes num(i)} $      num(i)为i在 gcd(i,j) (i <=n, j <= m)中出现的次数。

易得：

$num(i) = sumlimits_{i|d} {mu (frac{d}{i}) imes leftlfloor {frac{n}{d}} ight floor }  imes leftlfloor {frac{m}{d}} ight floor $

然后原式可以化为:

$sumlimits_{i = 1}^{min (n,m)} {F(i)}  imes sumlimits_{i|d} {mu (frac{d}{i}) imes leftlfloor {frac{n}{d}} ight floor }  imes leftlfloor {frac{m}{d}} ight floor$

换元:

$sumlimits_{d = 1}^{min (n,m)} {leftlfloor {frac{m}{d}} ight floor leftlfloor {frac{m}{d}} ight floor } sumlimits_{i|d} {F(i)mu (frac{d}{i})} $

然后再把$sumlimits_{i|d} {F(i)mu (frac{d}{i})} $预处理出来,对询问以a为关键字排序,用树状数组记录一段的和，前面用分块，然后就可以做了。

/*
唐代杜牧的《遣怀》
落魄江南载酒行，楚腰纤细掌中轻。
十年一觉扬州梦，赢得青楼薄幸名。 
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <iostream>
#include <string>
#include <ctime>
#include <map>
#include <set> 
#define LOCAL
long long MOD = 1000000000 + 7;
const int MAXM = 1000 * 1000 + 10;
const int MAXN = 100000 + 10;
using namespace std;
//输入输出优化 
int read(){
    int x = 0, flag = 1;
    char ch = getchar();
    while (ch < '1' || ch >'9') {if (ch == '-') flag = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = x * 10 + (ch - '0'); ch = getchar();}
    return x * flag;
}
struct FF{
    int order;//order表示该num对应的gcd值
    int num;
    bool operator < (const FF &b)const{
        return num < b.num;
    }
}F[MAXN];
struct QUERY{
    int l, r, a, order;
    bool operator < (const QUERY &b)const{
        return a < b.a;
    }
}q[MAXN];
int mu[MAXN], prime[MAXN];
int g[MAXN], C[MAXN], Q, Ans[MAXN];

int lowbit(int x){return x & -x;}
void add(int x, int val){
    while (x <= 100000){
        C[x] += val;
        x += lowbit(x);
    }
    return;
}
int sum(int x){
    int cnt = 0;
    while (x > 0){
        cnt += C[x];
        x -= lowbit(x);
    }
    return cnt;
}
void prepare(){
    memset(prime, 0, sizeof(prime));
    memset(C, 0, sizeof(C));
    
    mu[1] = 1;
    for (int i = 2;i <= 100000; i++){
        if (!prime[i]){
            prime[++prime[0]] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= prime[0]; j++){
            if (i * prime[j] > 100000) break;
            prime[i * prime[j]] = 1;
            if (i % prime[j] == 0){
                mu[i * prime[j]] = 0;
                break;
            }else mu[i * prime[j]] = -mu[i];
        }
    }
    //F[i]代表i的因数和
    F[1].num = F[1].order = 1;
    for (int i = 2; i <= 100000; i++){
        int cnt = 0;
        for (long long j = 1; j * (long long)j <= (long long)i; j++){
            if (j * j == i){cnt += j; break;}
            if (i % j != 0) continue;
            cnt += j + (i / j);
        }
        F[i].num = cnt;
        F[i].order = i;
    }
    sort(F + 1, F + 1 + 100000);
    //for (int i = 1; i <= 100; i++) printf("%d
", mu[i]);
}
void init(){
    scanf("%d", &Q);
    for (int i = 1; i <= Q; i++){
        int l = read(), r = read(), a = read();
        q[i].l = l; q[i].r = r;
        q[i].a = a; q[i].order = i;
    }
    sort(q + 1, q + 1 + Q);
}
//直接回答第x个询问
int query(int x){
    int cnt = 0;
    for (int i = 1; i <= min(q[x].l, q[x].r); i++){
        int t = min(q[x].l / (q[x].l / i), q[x].r / (q[x].r / i));
        cnt += (q[x].l / i) * (q[x].r / i) * (sum(t) - sum(i - 1));
        i = t;
    }
    return cnt;
}
void work(){
    int pos = 1;//表示a现在的大小
    for (int i = 1; i <= Q; i++){
        while (F[pos].num <= q[i].a && pos <= 100000){
            for (int j = 1; j * F[pos].order <= 100000; j++)
                add(j * F[pos].order, F[pos].num * mu[j]);
            pos++;
        }
        Ans[q[i].order] = query(i);
    }
    for (int i = 1; i <= Q; i++) printf("%d
", Ans[i] & 0x7fffffff);
}

int main(){
    int T;

     prepare();
     init();
     work();
     return 0;
}
 

 

(∑k=1nakbk)2≤(∑k=1na2k)(∑k=1nb2k)