codeforces 691F 暴力

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

题意:给你n个数和q次询问,每次询问问你有多少对ai,aj满足ai*aj>=q[i],注意 a*b 与 b*a是不同的

题解:sum[i]记录的是两个数乘积为i的方法数,然后前缀和记录小于等于乘积为i的方法个数,输出答案就容斥一下,因为n个数最多组成n*(n-1)/2对数,减去小于乘积为q[i]的数后即为乘积大于等于q[i]的方法个数

为什么可以暴力是因为注意到了值域的范围为3e6,调和级数的复杂度为nlogn,O(能过)

代码:

/**
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 *        ┃       ┃  
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 *        ┃       ┃
 *        ┃... ⌒ ...  ┃
 *        ┃       ┃
 *        ┗━┓   ┏━┛
 *          ┃   ┃ Code is far away from bug with the animal protecting          
 *          ┃   ┃   神兽保佑,代码无bug
 *          ┃   ┃           
 *          ┃   ┃        
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 *          ┃   ┃           
 *          ┃   ┗━━━┓
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 *           ┃┫┫ ┃┫┫
 *           ┗┻┛ ┗┻┛
 */
// warm heart, wagging tail,and a smile just for you!
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//                         ███████                           █████  ███████████████████
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 3e6 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int a[maxn];
LL num[maxn];
LL sum[maxn];
int p[maxn];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n;
    int maxx = -1;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        num[a[i]]++;
    }

    for(int i = 0; i <= 3000000; i++) {
        for(int j = 0; j <= 3000000; j++) {
            if(i * j > 3000000) break;
            if(i == j) sum[i * j] -= num[i];
            sum[i * j] += num[i] * num[j];
        }
    }
    for(int i = 1; i <= 3000000; i++) {
        sum[i] += sum[i - 1];
    }
    int m;
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        scanf("%d", &p[i]);
        printf("%lld\n", (LL)(n - 1)*n - sum[p[i] - 1]);
    }

    return 0;
}
View Code

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转载自www.cnblogs.com/buerdepepeqi/p/10662994.html