HDUOJ 4282 A very hard mathematic problem

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <sstream>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <cctype>#define CLOSE() ios::sync_with_stdio(false)#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)const int maxn = 46500;using LL = long long;using UI = unsigned int;using namespace std;//------------------------------------------------------------------------------------------////暴力枚举，注意z = 2时为完全平方，单独处理，避免tleusing ULL = unsigned long long;ULL a[maxn][31];ULL qpow(ULL a, ULL b) {ULL ret = 1;while (b) {if (b & 1) ret = ret * a;a = a * a;b >>= 1;}return ret;}int main() {#ifdef _DEBUGIN(); OUT();#endifULL k;ULL x, y, z;while (scanf("%lld", &k) && k) {int cnt = 0;ULL t = (ULL)sqrt(k);if (t * t == k) cnt += (t - 1) / 2;for (z = 3; z <= 30; ++z) {for (x = 1; ; ++x) {ULL xz = qpow(x, z);if (xz * 2 >= k) break;for (y = x + 1; ; y++) {ULL yz = qpow(y, z);if (xz + yz + x* y* z > k) break;if (xz + yz + x * y * z == k) {cnt++;printf("%lld = %lld^%lld + %lld^%lld + %lld * %lld * %lld\n", k, x, z, y, z, x, y, z);}}}}cout << cnt << endl;}return 0;}

#include <cstdio>#include <cmath>#define OUT() freopen("out.txt", "w", stdout);#define IN() freopen("in.txt", "r", stdin);using namespace std;const int maxn = 46500;using ULL = unsigned long long;//二分法，枚举y, z二分x。ULL a[maxn][31];ULL qpow(ULL a, ULL b) {ULL ret = 1;while (b) {if (b & 1) ret = ret * a;a = a * a;b >>= 1;}return ret;}void fun() {for (ULL i = 1; i < maxn; ++i) {for (ULL j = 2; j <= 30; ++j) {ULL t = qpow(i, j);if (t < 2147483648) {a[i][j] = t;//cout << t << endl;}}}}int main() {#ifdef _DEBUGIN(); OUT();#endiffun();ULL k;ULL x, y, z;while (scanf("%lld", &k) && k) {int cnt = 0;for (z = 2; z <= 30; ++z) {int t = sqrt(k) + 1;for (y = 2; y <= t; ++y) {if (!a[y][z]) break;if (a[y][z] + y*z >= k) break;ULL l = 1, r = y-1;while (l <= r) {x = (l + r) / 2;ULL t = a[x][z] + a[y][z] + x*y*z;if (t > k) r = x - 1;else if (t < k) l = x + 1;else {cnt++;//printf("%lld = %lld^%lld + %lld^%lld + %lld * %lld * %lld\n", k, x, z, y, z, x, y, z);break;}}}}printf("%d\n", cnt);}return 0;}