[51nod] 1024 矩阵中不重复的元素

这个最暴力的想法就是暴力模拟算出详细值加set去重

时间复杂度空间复杂度过大。

所以需要对值进行映射，再用set去重

映射方式很多，比较方便的是取对数

logn(a ^ b) = b * logn(a)

n可以取

e->log()

10->log10()

2->log2()

因为取了对数肯定会有误差

所以需要重载一下EPS

或者log2的精度比较高 可以不重载

/*
    Zeolim - An AC a day keeps the bug away
*/

//#pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <sstream>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef long long ll;

const int MAXN = 1e6 + 10;

const double eps = 0.0000001;

struct num
{
    double val;

    num(){}

    num(double x) 
    {
        val = x;
    }

    bool operator <(const num &b)  const
    {
        return val + eps < b.val;
    }

    bool operator == (const num &b) const
    {
        return (val - b.val) <= eps;
    }
};

int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);     cout.tie(0);
    //freopen("D://test.in", "r", stdin);
    //freopen("D://test.out", "w", stdout);

    set <num> S;

    int m, n, a, b;

    cin>>m>>n>>a>>b;
    

    for(int i = 0; i < n; i++) 
    {
        for(int j = 0; j < m; j++) 
        {
            S.insert( num ( double(j + b) * log10( double(a + i) ) ) );
        }
    }

    cout<<S.size()<<endl;

    return 0;
}