Project Euler：Problem 61 Cyclical figurate numbers

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle P_3,n=n(n+1)/2 1, 3, 6, 10, 15, ...Square P_4,n=n² 1, 4, 9, 16, 25, ...Pentagonal P_5,n=n(3n−1)/2 1, 5, 12, 22, 35, ...Hexagonal P_6,n=n(2n−1) 1, 6, 15, 28, 45, ...Heptagonal P_7,n=n(5n−3)/2 1, 7, 18, 34, 55, ...Octagonal P_8,n=n(3n−2) 1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P_3,127=8128), square (P_4,91=8281), and pentagonal (P_5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

又暴力破解了一次ㄟ( ▔, ▔ )ㄏ

一开始没看清题意，我以为这些数依次是满足triangle, square, pentagonal, hexagonal, heptagonal, and octagonal，结果发现无解┑(￣Д ￣)┍

#include <iostream>#include <string>#include <vector>#include <unordered_map>#include <time.h>using namespace std;int triangle[100];int pentagonal[10000];int hextagonal[10000];int heptagonal[10000];int octagonal[10000];int tri_count = 0;void getTriangle(){int count = 0;for (int i = 1; i <= 200; i++){int num = i*(i + 1) / 2;if (num >1000&&num<10000)triangle[count++] = num;}tri_count = count;}bool isSqure(int n){int i = sqrt(n);if (i*i == n&&n>1000&&n<10000)return true;return false;}void getPentagonal(){for (int i = 1; i <= 200; i++){int num = i*(3 * i - 1) / 2;if (num > 1000 && num < 10000)pentagonal[num] = 1;}}bool isPentagonal(int n){if (pentagonal[n] == 1)return true;return false;}void getHexagonal(){for (int i = 1; i <= 200; i++){int num = i*(2 * i - 1);if (num>1000 && num < 10000)hextagonal[num] = 1;}}bool isHexagonal(int n){if (hextagonal[n] == 1)return true;return false;}void getHeptagonal(){for (int i = 1; i <= 200; i++){int num = i*(5 * i - 3) / 2;if (num > 1000 && num < 10000)heptagonal[num] = 1;}}bool isHeptagonal(int n){if (heptagonal[n] == 1)return true;return false;}void getOctagonal(){for (int i = 1; i <= 200; i++){int num = i*(3 * i - 2);if (num > 1000 && num < 10000)octagonal[num] = 1;}}bool isOctagonal(int n){if (octagonal[n] == 1)return true;return false;}bool(*figurate[5])(int) = { isSqure, isPentagonal, isHexagonal, isHeptagonal, isOctagonal };vector<int> GetRandomSequence() {unordered_map<int, int>tab;vector<int>res;int num;for (int i = 0; i < 5; i++){do{num = rand() % 5;} while (tab.find(num) != tab.end());tab.insert(make_pair(num, 1));res.push_back(num);}return res;}int check(){int sum = 0;srand((int)time(0));vector<int>rs = GetRandomSequence();for (int i = 0; i < tri_count; i++){int a = triangle[i] / 100;int b = triangle[i] % 100;for (int s = 10; s <= 99; s++){if ((*figurate[rs[0]])(b * 100 + s)){for (int p = 10; p <= 99; p++){if ((*figurate[rs[1]])(s * 100 + p)){for (int hx = 10; hx <= 99; hx++){if ((*figurate[rs[2]])(p * 100 + hx)){for (int hp = 10; hp <= 99; hp++){if ((*figurate[rs[3]])(hx * 100 + hp)){if ((*figurate[rs[4]])(hp * 100 + a)){sum = triangle[i] + b * 100 + s + s * 100 + p + p * 100 + hx + hx * 100 + hp + hp * 100 + a;return sum;}}}}}}}}}}return -1;}int main(){memset(pentagonal, 0, sizeof(pentagonal));memset(hextagonal, 0, sizeof(hextagonal));memset(heptagonal, 0, sizeof(heptagonal));memset(octagonal, 0, sizeof(octagonal));getTriangle();getPentagonal();getHexagonal();getHeptagonal();getOctagonal();int flag;while (true){flag = check();if (flag != -1)break;}cout << flag << endl;system("pause");return 0;}

把那个随机生成全排列换成next_permutation也是能搞出来的。