HDU 4676 Sum Of Gcd

Sum Of Gcd

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 405    Accepted Submission(s): 194

做了2013杭州的H题后回头看了下这题，发现之前根本不懂- -!

这题也是离线，好想要分块，就是那个sqrt(1.0*n)和lx[i].c=lx[i].l/d;和cmp

其他的用了个公式sigama(num[i],2)*euler[i];

公式中的num意义为能整除每个因子的数的个数

num[i]，l to r 中i能整除i的个数，euler[i]为i的欧拉函数！

后面的是指num[vt[a[i]][j]]，也就是能整除vt[a[i]][j]的个数和euler[vt[a[i]][j]];

这其实把num[i]*(num[i]-1)/2;这个等差求和的公式化回去，就相当于从0开始一个一个加

这题也要预处理因子，有那个因子就num[v]++;删的时候就num[v]--；便于整体相加求和！，然后按l排序，从左到右加过去，进行删补就可以了！

Problem Description

Given you a sequence of number a₁, a₂, ..., a_n, which is a permutation of 1...n.
You need to answer some queries, each with the following format:
Give you two numbers L, R, you should calculate sum of gcd(a[i], a[j]) for every L <= i < j <= R.

 

 

Input

First line contains a number T(T <= 10),denote the number of test cases.
Then follow T test cases.
For each test cases,the first line contains a number n(1<=n<= 20000).
The second line contains n number a₁,a₂,...,a_n.
The third line contains a number Q(1<=Q<=20000) denoting the number of queries.
Then Q lines follows,each lines contains two integer L,R(1<=L<=R<=n),denote a query.

 

 

Output

For each case, first you should print "Case #x:", where x indicates the case number between 1 and T.
Then for each query print the answer in one line.

 

 

Sample Input

1

5

3 2 5 4 1

3

1 5

2 4

3 3

 

 

Sample Output

Case #1:

11

4

0

 

 

Source

2013 Multi-University Training Contest 8

 

 

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#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 20005
#define ll long long
#define INF 0x7fffffff
#define eps 1e-8
struct line{int l, r,p,c;};
int n, m, k,x,y,res;
vector<int>p[maxn];
line lx[maxn];
int euler[maxn];
int num[maxn];
int ans[maxn];
int a[maxn];
void init(){
    for (int i = 1; i < maxn; i++)euler[i] = i;
    for (int i = 2; i < maxn; i++)
    if (euler[i] == i)
    for (int j = i; j < maxn; j += i)
        euler[j] = euler[j] / i*(i - 1);
    for (int i = 1; i < maxn; i++){
        p[i].clear();
        for (int j = 1; j*j <= i; j++)
        if (i%j == 0){
            p[i].push_back(j);
            if (j*j!=i)p[i].push_back(i / j);
        }
    }
}
int query(int l, int r){
    for (int i = l; i < x; i++)
    for (int j = 0; j < p[a[i]].size(); j++){
        k = p[a[i]][j];
        res += euler[k] * num[k];
        num[k]++;
    }
    for (int i = x; i < l; i++)
    for (int j = 0; j < p[a[i]].size(); j++){
        k = p[a[i]][j];
        num[k]--;
        res -= euler[k] * num[k];
    }
    for (int i = r+1; i <= y; i++)
    for (int j = 0; j < p[a[i]].size(); j++){
        k = p[a[i]][j];
        num[k]--;
        res -= euler[k] * num[k];
    }
    for (int i = y+1; i <= r; i++)
    for (int j = 0; j < p[a[i]].size(); j++){
        k = p[a[i]][j];
        res += euler[k] * num[k];
        num[k]++;
    }
    x = l; y = r;
    return res;
}
int cmp(line a, line b){
    if (a.c == b.c)return a.r < b.r;
    return a.c < b.c;
}
int main(){
    int cas = 1;
    init();
    int t;
    scanf("%d", &t);
    while (t--){
        scanf("%d", &n);
        memset(num, 0, sizeof num);
        for (int i = 1; i <= n; i++)scanf("%d", &a[i]);
        scanf("%d", &m);
        int d = sqrt(1.0*n);
        for (int i = 0; i < m; i++){
            scanf("%d%d", &lx[i].l, &lx[i].r);
            lx[i].c = lx[i].l / d;
            lx[i].p = i;
        }
        sort(lx, lx + m, cmp);
        res = 0;
        x = lx[0].l; y = lx[0].r;
        for (int i = x; i <= y; i++)
        for (int j = 0; j < p[a[i]].size(); j++){
            k = p[a[i]][j];
            res += euler[k] * num[k];
            num[k]++;
        }
        ans[lx[0].p] = res;
        for (int i = 1; i < m; i++)ans[lx[i].p] = query(lx[i].l, lx[i].r);
        printf("Case #%d:\n", cas++);
        for (int i = 0; i < m;i++)printf("%d\n", ans[i]);
    }
    return 0;
}