UVa Problem 10177 (2/3/4)-D Sqr/Rects/Cubes/Boxes? （2/3/4-维立方体？）

// (2/3/4)-D Sqr/Rects/Cubes/Boxes? （2/3/4-维立方体？）// PC/UVa IDs: 111206/10177, Popularity: B, Success rate: high Level: 2// Verdict: Accepted// Submission Date: 2011-11-01// UVa Run Time: 0.016s//// 版权所有（C）2011，邱秋。metaphysis # yeah dot net//// [解题方法]// 设边长为 n，则有以下结果：// 2 维：正方形数量 S2 形成的数列后项与前项差为 n^2。// 3 维：立方体数量 S3 形成的数列后项与前项差为 n^3。// 4 维：超立方体数量 S4 形成的数列后项与前项差为 n^4。//// 2 维：长方形数量 R2 = S3 - S2。// 3 维：长方体数量 R3 = S2 * S1，其中 S1 = (n - 1) * (n + 2) / 2。// 4 维：超长方体数量 R4 = (n * (n + 1) / 2)^4 - S4。#include <iostream>#include <cmath>using namespace std;#define MAXN 101#define MAXD 6#define S2 0#define R2 1#define S3 2#define R3 3#define S4 4#define R4 5long long count[MAXN][MAXD];int main(int ac, char *av[]){for (int n = 1; n < MAXN; n++){count[n][S2] = count[n - 1][S2] + pow(n, 2);count[n][S3] = count[n - 1][S3] + pow(n, 3);count[n][S4] = count[n - 1][S4] + pow(n, 4);count[n][R2] = count[n][S3] - count[n][S2];count[n][R3] = count[n][S3] * (n - 1) * (n + 2) / 2;count[n][R4] = pow(n * (n + 1) / 2, 4) - count[n][S4];}int size;while (cin >> size){for (int d = 0; d < MAXD; d++)cout << (d ? " " : "") << count[size][d];cout << endl;}return 0;}