【codeforces 691 F】【数论+前缀和 好题】【求序列中乘积大于等于p的点对的个数】

传送门:http://codeforces.com/contest/691/problem/F

题意：求序列中乘积大于等于p的点对的个数

刚开始还想用数据结构维护一下，想想不靠谱

先反向思考求乘积小于p的情况，具体的有两种思路

思路一：类似于埃式筛法，cnt[a[i]]记录a[i]的个数，mul[i]记录点对乘积是i*j的个数

复杂度：O（NloglogN+m）

代码：

#include <bits/stdc++.h>
using namespace std;

#define ll __int64
const int N=3e6+10;

int cnt
, a
, p
;
ll mul
;

int main(){
int n, m;
scanf("%d", &n);
for(int i=1; i<=n; i++){
scanf("%d", &a[i]);
cnt[a[i]]++;
}
scanf("%d", &m);
for(int i=1; i<=m; i++)scanf("%d", &p[i]);

for(int i=1; i<N; i++){
if(!cnt[i])continue;
for(int j=i; j<N; j+=i){//用埃式筛法的思想处理乘积问题
mul[j]+=1LL*cnt[i]*(cnt[j/i]-(j/i == i));
/* if(j/i!=i)mul[j]+=1LL*cnt[i]*cnt[j/i];
else mul[j]+=1LL*cnt[i]*(cnt[i]-1); //mul[j]表示乘积为j的个数*/
}
}

for(int i=1; i<N; i++)mul[i]+=mul[i-1];

for(int i=1; i<=m; i++){
ll ans=1LL*n*(n-1)-mul[p[i]-1]; //不要忘记LL
printf("%I64d\n", ans);
}
return 0;
}

思路二：
先去重再离散化，这时候a[i]表示的下标，按下标枚举去重后的数组并且点对的乘积小于p，这样子均摊的复杂度还是O（N）

代码：

#include <bits/stdc++.h>
using namespace std;

#define ll __int64
const int N=3e6+10;

int cnt
, a
, b
;
ll mul
;

int main(){
int n, m;
scanf("%d", &n);
for(int i=1; i<=n; i++){
scanf("%d", &a[i]);
b[i]=a[i];
}
scanf("%d", &m);

sort(b+1, b+n+1);
int tol=unique(b+1, b+n+1)-(b+1);//离散化
for(int i=1; i<=n; i++)
a[i]=lower_bound(b+1, b+tol+1, a[i])-b;

for(int i=1; i<=n; i++)cnt[a[i]]++;
ll all=1LL*n*(n-1);
n=tol;

for(int i=1; i<=n; i++){//复杂度大概O(sqrt(n)*sqrt(n))
for(int j=1; j<=n; j++){
if(b[i]*b[j]<N) mul[b[i]*b[j]]+=cnt[i]*(cnt[j]-(i==j));
else break;
}
}

for(int i=1; i<N; i++)mul[i]+=mul[i-1];

while(m--){
int p;
scanf("%d", &p);
printf("%I64d\n", all-mul[p-1]);
}
return 0;
}

F. Couple Cover

time limit per test
3 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Couple Cover, a wildly popular luck-based game, is about to begin! Two players must work together to construct a rectangle. A bag with n balls,
each with an integer written on it, is placed on the table. The first player reaches in and grabs a ball randomly (all balls have equal probability of being chosen) — the number written on this ball is the rectangle's width in meters. This ball is not returned
to the bag, and the second player reaches into the bag and grabs another ball — the number written on this ball is the rectangle's height in meters. If the area of the rectangle is greater than or equal some threshold p square
meters, the players win. Otherwise, they lose.

The organizers of the game are trying to select an appropriate value for p so that the probability of a couple winning is not too high
and not too low, but they are slow at counting, so they have hired you to answer some questions for them. You are given a list of the numbers written on the balls, the organizers would like to know how many winning pairs of balls exist for different values
of p. Note that two pairs are different if either the first or the second ball is different between the two in pair, and two different
balls with the same number are considered different.

Input

The input begins with a single positive integer n in its own line (1 ≤ n ≤ 106).

The second line contains n positive integers — the i-th
number in this line is equal to ai (1 ≤ ai ≤ 3·106),
the number written on the i-th ball.

The next line contains an integer m (1 ≤ m ≤ 106),
the number of questions you are being asked.

Then, the following line contains m positive integers — the j-th
number in this line is equal to the value of p(1 ≤ p ≤ 3·106)
in the j-th question you are being asked.

Output

For each question, print the number of winning pairs of balls that exist for the given value of p in the separate line.

Examples

input

5
4 2 6 1 3
4
1 3 5 8

output

20
18
14
10

input

2
5 6
2
30 31

output

2
0