UVA 10177 (2/3/4)-D Sqr/Rects/Cubes/Boxes? 简单数学题

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求一个大小是N的正方形中正方形的个数S2，长方形的个数R2;

求一个棱长为N的正方体中正方体的个数S3，长方体的个数R3;

求一个边长为N的超正方体中超正方体的个数S4，超长方体的个数R4；

在以上结果中正方形不视为长方形。。。

首先对于S2和R2，

显然S2的个数为n*n + (n - 1)*(n - 1) + (n - 2)*(n - 2) + ... + 1*1 即边长为k的正方形的个数为 (n + 1 - k) * (n + 1 - k) 个

对于R2首先考虑整个正方形中可以得到的矩形的个数，考虑矩形左上角对应的是原正方形中的哪个，而右下角的对应的又是哪个

则可以知道对于以第 i 行 j 列的正方形作为左上角的矩形有 (n + 1 - i)*(n + 1 - j) 种；

那么 R2 = sigma(sigma(i*j)) - sigma(i*i) = n*n*(n + 1)*(n + 1) / 4 - n * (n + 1) * (2*n + 1) / 6

类似的可以知道 S3 = sigma(i*i*i) = n*n*(n + 1)*(n + 1) / 4;

R3 = sigma(sigma(sigma(i*j*k))) - sigma(i*i*i) = (n*(n + 1)/2)^3 - n*n*(n + 1)*(n + 1) / 4;

S4 = sigma(i*i*i*i) = n * (n + 1) * (2*n + 1) * (3*n*n + 3*n - 1) / 30

R4 = sigma(sigma(sigma(sigma(i*j*k*r)))) - sigma(i*i*i*i) = (n * (n + 1) / 2)^4 - S4;

代码如下：

Result : Accepted Memory : 0 KB Time : 19 ms

/* * Author: Gatevin * Created Time:  2014/7/7 21:36:38 * File Name: test.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;int N;long long S2[101];long long R2[101];long long S3[101];long long R3[101];long long S4[101];long long R4[101];int main(){    memset(S2, 0, sizeof(S2));    memset(R2, 0, sizeof(R2));    memset(S3, 0, sizeof(S3));    memset(R3, 0, sizeof(R3));    memset(S4, 0, sizeof(S4));    memset(R4, 0, sizeof(R4));    for(long long i = 0; i <= 100; i++)    {        long long tmp = (i * (i + 1)) / 2;        S2[i] = (i * (i + 1) * (2*i + 1)) / 6;        S3[i] = tmp * tmp;        R2[i] = S3[i] - S2[i];        R3[i] = S3[i]*tmp - S3[i];        S4[i] = (i * (i + 1) * (2*i + 1) * (3 * i * i + 3 * i - 1)) / 30;        R4[i] = S3[i]*S3[i] - S4[i];    }    while(cin>>N)    {        cout<<S2[N]<<" "<<R2[N]<<" "<<S3[N]<<" "<<R3[N]<<" "<<S4[N]<<" "<<R4[N]<<endl;    }    return 0;}

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