Project Euler 45: Triangular, pentagonal, and hexagonal

题目描述:

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle		Tn=n(n+1)/2		1, 3, 6, 10, 15, ...
Pentagonal		Pn=n(3n−1)/2		1, 5, 12, 22, 35, ...
Hexagonal		Hn=n(2n−1)		1, 6, 15, 28, 45, ...

It can be verified that T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

 

这道题目的求解是按照下述方法进行的

如果一个数x是一个Triangle number，则(sqrt(8x+1)-1)/2为一个整数。

如果一个数x是一个pentagonal number，则(sqrt(24x+1)+1)/6是一个整数。

如果一个数x是一个hexagonal number，则(sqrt(8x+1)+1)/4是一个整数。

因此采用最暴力的办法便是从第144个hexagonal number开始不断地枚举并判断这个数是不是Triangle number和pentagonal number。但是进一步我们可以发现H(n)=T(2n-1)的，因此我们在判断的时候可以直接判断一个数同时是hexagonal number和pentagonal number即可。

 

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
#include <iostream>
#include <fstream>
#include <sstream>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <numeric>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll  INF = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-8;
const double pi  = acos(-1.0);

int main() {

    for(int i = 144; ; ++i) {
        long long t = i * (2 * i - 1);
        double t1 = (sqrt(24 * t * 1.0 + 1.0) + 1) / 6.0;
        long long t2 = (int)t1;
        if(abs(t1 - t2) < eps) {
            cout << t << endl;
            break;
        }
    }

    return 0;
}