Problem Statement

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Definition

Limits

Notes

• rectangles with equals sides (squares) should not be counted.

Constraints

• width and height will be between 1 and 1000 inclusive.

Test cases

0.  

   • width3
   • height3

    

   Returns22

    

   See above
1.  

   • width5
   • height2

    

   Returns31

    

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   In this grid, there is one 2x5 rectangle, 2 2x4 rectangles, 2 1x5 rectangles, 3 2x3 rectangles, 4 1x4 rectangles, 6 1x3 rectangles and 13 1x2 rectangles. Thus there is a total of 1 + 2 + 2 + 3 + 4 + 6 + 13 = 31 rectangles.

2.  

   • width10
   • height10

    

   Returns2640
3.  

   • width1
   • height1

    

   Returns0
4.  

   • width592
   • height964

    

   Returns81508708664
   

   

     1 #include <cstdio>
     2 #include <cmath>
     3 #include <cstring>
     4 #include <ctime>
     5 #include <iostream>
     6 #include <algorithm>
     7 #include <set>
     8 #include <vector>
     9 #include <sstream>
    10 #include <typeinfo>
    11 #include <fstream>
    12 
    13 using namespace std;
    14 typedef long long ll ;
    15 class RectangularGrid {
    16     public:
    17     long long countRectangles(int l , int r) {
    18         if (l > r) swap (l , r ) ;
    19       //  printf ("l = %d , r = %d\n" , l , r ) ;
    20         ll all = (1ll *l*l*r*r + 1ll *l*r*r + 1ll *r*l*l + 1ll *r*l) / 4 ;
    21         ll sum = 1ll * l * (l - 1) * (l + 1) / 3 + 1ll * (r - l + 1) * l * (l + 1) / 2 ;
    22      //   printf ("all = %lld , sum = %lld\n" , all , sum) ;
    23         return all - sum ;
    24     }
    25 };
    26 
    27 // CUT begin
    28 ifstream data("RectangularGrid.sample");
    29 
    30 string next_line() {
    31     string s;
    32     getline(data, s);
    33     return s;
    34 }
    35 
    36 template <typename T> void from_stream(T &t) {
    37     stringstream ss(next_line());
    38     ss >> t;
    39 }
    40 
    41 void from_stream(string &s) {
    42     s = next_line();
    43 }
    44 
    45 template <typename T>
    46 string to_string(T t) {
    47     stringstream s;
    48     s << t;
    49     return s.str();
    50 }
    51 
    52 string to_string(string t) {
    53     return "\"" + t + "\"";
    54 }
    55 
    56 bool do_test(int width, int height, long long __expected) {
    57     time_t startClock = clock();
    58     RectangularGrid *instance = new RectangularGrid();
    59     long long __result = instance->countRectangles(width, height);
    60     double elapsed = (double)(clock() - startClock) / CLOCKS_PER_SEC;
    61     delete instance;
    62 
    63     if (__result == __expected) {
    64         cout << "PASSED!" << " (" << elapsed << " seconds)" << endl;
    65         return true;
    66     }
    67     else {
    68         cout << "FAILED!" << " (" << elapsed << " seconds)" << endl;
    69         cout << "           Expected: " << to_string(__expected) << endl;
    70         cout << "           Received: " << to_string(__result) << endl;
    71         return false;
    72     }
    73 }
    74 
    75 int run_test(bool mainProcess, const set<int> &case_set, const string command) {
    76     int cases = 0, passed = 0;
    77     while (true) {
    78         if (next_line().find("--") != 0)
    79             break;
    80         int width;
    81         from_stream(width);
    82         int height;
    83         from_stream(height);
    84         next_line();
    85         long long __answer;
    86         from_stream(__answer);
    87 
    88         cases++;
    89         if (case_set.size() > 0 && case_set.find(cases - 1) == case_set.end())
    90             continue;
    91 
    92         cout << "  Testcase #" << cases - 1 << " ... ";
    93         if ( do_test(width, height, __answer)) {
    94             passed++;
    95         }
    96     }
    97     if (mainProcess) {
    98         cout << endl << "Passed : " << passed << "/" << cases << " cases" << endl;
    99         int T = time(NULL) - 1433920510;
   100         double PT = T / 60.0, TT = 75.0;
   101         cout << "Time   : " << T / 60 << " minutes " << T % 60 << " secs" << endl;
   102         cout << "Score  : " << 500 * (0.3 + (0.7 * TT * TT) / (10.0 * PT * PT + TT * TT)) << " points" << endl;
   103     }
   104     return 0;
   105 }
   106 
   107 int main(int argc, char *argv[]) {
   108     cout.setf(ios::fixed, ios::floatfield);
   109     cout.precision(2);
   110     set<int> cases;
   111     bool mainProcess = true;
   112     for (int i = 1; i < argc; ++i) {
   113         if ( string(argv[i]) == "-") {
   114             mainProcess = false;
   115         } else {
   116             cases.insert(atoi(argv[i]));
   117         }
   118     }
   119     if (mainProcess) {
   120         cout << "RectangularGrid (500 Points)" << endl << endl;
   121     }
   122     return run_test(mainProcess, cases, argv[0]);
   123 }
   124 // CUT end

   View Code

   已知一个长 l , 宽 r 的由１×１的单位矩阵构成的矩阵，问其中有多少个长方形？

   那么我们只要求出其中矩阵的总个数n , 和正方形数m,然后n - m就是答案。

   ｎ＝ ｌ*ｒ　＋　ｌ×（ｒ×（ｒ－１））／２　＋　ｒ×（ｌ×（ｌ－１））／２　＋　ｌ×ｒ×（ｌ－１）×（ｒ－１）／４

   　＝（ｌ×ｌ×ｒ×ｒ＋ｌ×ｒ×ｒ＋ｒ×ｌ×ｌ＋ｒ×ｌ）／４；

   ｍ＝２×（１＋２×３／２＋４×３／２＋……＋ｌ×（ｌ－１）／２）　＋　（ｒ－ｌ＋１）×ｌ×（ｌ＋１）／２；

   　＝ｌ×（ｌ＋１）×（２×ｌ＋１）／６　＋　（ｒ－ｌ）×ｌ×（ｌ＋１）／２；

   ＰＳ：

   1^2 + 2^2 + 3^2 + 4^2 + 5^2 +……＋n^2 = n*(n + 1)*(2*n+1)/6;

   证明：（来自百度文库）

   想像一个有圆圈构成的正三角形，

   第一行1个圈，圈内的数字为1  

   第二行2个圈，圈内的数字都为2，

   以此类推

   第n行n个圈，圈内的数字都为n，

   我们要求的平方和，

   就转化为了求这个三角形所有圈内数字的和。

   设这个数为r  

   下面将这个三角形顺时针旋转120度，

   得到第二个三角形 

   再将第二个三角形顺时针旋转120度，得到第三个三角形.

    然后，将这三个三角形对应的圆圈内的数字相加，

   我们神奇的发现所有圈内的数字都变成了

   2n+1  

    而总共有几个圈呢，这是一个简单的等差数列求和

     1+2+……+n=n(n+1)/2 

    于是

   3r=[n(n+1)/2]*(2n+1)  

   r=n(n+1)(2n+1)/6