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#include <iostream>using namespace std;/** The main idea of the problem: * 1, average bar this 2 of the K power of the squares divided into 4 parts * 2, a special box containing a part of * 3, There are 3 left without special squares to contribute a small block to synthesize an "L", and then make the small square that you just contributed to as a special square. Again back to step 1*5, assuming that the special box is less than 4 is already all divided, jump out *///we have to know is K row k column square. So there will be K, know the position of the special squares row and column th, TL, the starting position of the new squares after dividing newh,newl, rows and columns void Qipan (int k, int th, int tl, int newh, int newl, int a[]) {//Start cutting chess --------------------------------*/int t=0;//dominoes number int s=k/2;//cut checkerboard//int a[4] = {0};//represents four different L-shaped squares//At the beginning of the division of 4 parts, Special squares can exist in 4 chunks of each block//special squares in the upper left part if (Th < newh+s && tl < newl+s) {//Here special squares are at the top left, call 4th L Domino A[3]+=1;qipan (s, Th, TL, NEWH, NEWL, a);//S: This is the number of rows of the cutting board th: The high tl of special squares: The column of Special Squares NEWH: The starting line of this large square NEWL: The starting column}else{//there is no special squares in the upper left-hand board// So the lower left corner of the square as a part of L block Qipan (S, newh+s-1, Newl+s-1, NEWH, NEWL, a);} ------------------------------------------------------------------------------------//Special squares in the lower left part if (Th > NEWH +s-1 && TL < newl+s) {//Here the special squares are in the upper left., call No. 2nd L Domino a[1]++;qipan (s, Th, TL, newh+s, NEWL, a);//S: This is the number of rows of the cutting board th: special squares of high TL: special squares of columns NEWH: The starting row of this large square NEWL: Start column}else{// There is no special box in the upper left of the chessboard//that the lower left corner of the square as a part of L block Qipan (S, NEWH + S, newl + s-1, Newh+s, NEWL, a);} ------------------------------------------------------------------------------------/* Dominoes can have 4 kinds of-------------------------- ---------* * * * * * * * * * * * * * *------------------------------------*///Special squares on upper right if (Th < newh+s && tl > Newl+s-1) {//Here special squares are at the top left. Just call 3rd L Domino A[2]++;qipan (s, Th, TL, NEWH, Newl+s, a);//S: This is the line of the cutting board th: special squares of high TL: special squares of columns NEWH: The starting row of this large square NEWL: Start column}else{// There is no special box in the upper left of the chessboard//that the lower left corner of the square as a part of L block Qipan (s, Newh + s-1, NEWL + S, NEWH, Newl+s, a);} ------------------------------------------------------------------------------------//Special squares in the lower right part if (Th > NEWH +s-1 && tl > Newl+s-1) {//Here the Special square is at the top left, call 1th L Domino a[0]++;qipan (s, Th, TL, newh+s, NEWL + S, a);//S: This is the number of rows of the cutting board th: Special The high TL: special squares of the column NEWH: The starting line of this large square NEWL: The starting column}else{//the upper left of the board there is no special box//that the lower left corner of the square as a part of L with L block Qipan (s, NEWH +S, Newl + S, newh+s, Newl+s, a);}} /*void Test (int *b) {for (int i = 0; I <= 3; ++i) {b[i]++;if (i = = 2) {B[i] = 998;}}} */int Main () {char C;int a[4] = {0};qipan (4, 1, 2, 1, 1, a);//cout << "Altogether:" << Qipan (4, 1, 2, 1, 1, a) <&L t; " << endl;/* dominoes can have 4 kinds of-----------------------------------* * * * * * * * * * * * * * * * * Char L[4][2][2] =//uses a multidimensional array to represent these L-shaped dominoes {{{' * ', ' * '}, {' * ', '}}, {' * ', ' * '}, {', ' * '}},{{' * ', '}, {' * ', ' * '}},{{', ' * '}, {' * ', ' * '}}};for (int i = 0; i < 4; ++i) {for (int j = 0; j < 2; ++j)//First line of first Domino {for (i NT k = 0; K < 2; ++K)//The case of each line {cout << l[i][j][k];} cout << Endl;} cout << "<< i +1 <<" L-type dominoes are used in the following times: "<< A[i] << Endl;} /*test (a); for (int i = 0; i < 4; ++i) {cout << a[i] << Endl;} */cin >> C;return 0;}
q= design &t=blog "target=" _blank "> Design /
Q= Board Overlay &t=blog "target=" _blank "> Board overlay /
The general description of the problem is this: as you can see, (this is my copy on the Internet), this is a 4*4-sized chessboard.
For this kind of problem. The size of the chessboard is generally 2 ...
However, many others tend to "prove", which is what I and a lot of readers want to know, not the beginning of the code. Folder 0, description of the problem ... 11. Related Blogs ... 1 ...
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