Beauty of Programming

1.1 CPU usage Problems

# Include <iostream> # include <ctime> # include <cmath> # include <windows. h> using namespace STD; // method 1 void main () {int64 start = 0; int busy = 10; int idle = busy; cout <"CPU usage problem "; while (true) {start = gettickcount (); While (gettickcount ()-Start) <= busy); sleep (idle );}} // method 2 int main () {for (;) {for (INT I = 0; I <9600000; I ++ ); // For (INT I = 0; I <21360000; I ++); // 2.67 GHz 4-core sleep (10);} return 0 ;} // const Double Split = 0.01; const int COUNT = 200; const double Pi = 3.14159265; const int interval = 300; void main () {DWORD busy [count], idle [count]; int half = interval/2; double radian = 0.0; For (INT I = 0; I <count; I ++) {busy [I] = DWORD (sin (pI * radian) * Half + half); idle [I] = interval-busy [I]; radian ++ = 0.01 ;} DWORD start = 0; Int J = 0; while (true) {start = gettickcount (); j = J % count; while (gettickcount ()-start) <= busy [J]); sleep (idle [J]); j ++ ;}}

Number of CPU core running cycles

# Include <iostream> using namespace STD; inline _ int64 getcputickcount () {__ ASM {rdtsc ;}} void main () {cout <"Number of CPU core run cycles" <getcputickcount () <Endl; System ("pause ");}

1.2 generals

# Include <iostream> using namespace STD; // first method struct {unsigned char A: 4; unsigned char B: 4;} I; void main () {for (I. a = 1; I. A <= 9; I. A ++) for (I. B = 1; I. B <= 9; I. B ++) if (I. A % 3! = I. B % 3) printf ("A = % d, B = % d \ n", I. a, I. b); System ("pause");} // method 2 # define half_bits_length 4 // The value is half the length of the memory unit, in this question, it is 4bit # define fullmask 255 // This number represents a full-bit mask. In binary representation, it is 11111111. # Define lmask (fullmask 

1.12 elevator Scheduling

#include <iostream>using namespace std;#define N 6void main(){int nPerson[N]={55,66,77,88,99,44};int N1=0,N2=0,N3=0;int nTargetFloor=0,nMinFloor=0,i;for (i=1,N1=0,N2=nPerson[0],N3=0;i<N;i++){N3+=nPerson[i];nMinFloor+=nPerson[i+1]*i;}for (i=1;i<N;i++){if (N1+N2<N3){nTargetFloor=i+1;nMinFloor+=(N1+N2-N3);N1+=N2;N2=nPerson[i];N3-=nPerson[i];}elsebreak;}cout<<"nTargetFloor "<<nTargetFloor<<"\nnMinFloor "<<nMinFloor<<endl;system("pause");}

1.13 Nim two piles of stones

# Include <iostream> # include <cmath> using namespace STD; # define swap (x, y) (x) ^ = (y), (y) ^ = (x), (x) ^ = (y) void main () {double A, B; A = (1 + SQRT (5.0)/2; B = (3 + SQRT (5.0)/2; int M, N; bool Nim = false; cout <"number of books that input two piles of stones \ n "; cin> m> N; If (M = N) Nim = true; If (n> m) Swap (n, m ); if (n-M = (long) floor (N * A) Nim = false; else Nim = true; If (NIM) cout <"winning stone players first \ n"; elsecout <"winning stone players first \ n"; System ("pause ");}

2.7 Max public approx. Least public multiple
Http://blog.csdn.net/aoxiangzhiguanjun/article/details/8755260

2.13 Maximum Product of sub-array

# Include <iostream> # include <stdlib. h> # include <stdio. h> using namespace STD; // Maximum Product of the sub-array int maxproduct (int * a, int N) {int maxproduct = 1; // Max positive product at current position int minproduct = 1; // min negative product at current position int r = 1; // result, Max multiplication totally for (INT I = 0; I <n; I ++) {if (a [I]> 0) {maxproduct * = A [I]; minproduct = min (minproduct * A [I], 1);} else if (a [I] = 0) {maxproduct = 1; minproduct = 1;} else // A [I] <0 {int temp = maxproduct; maxproduct = max (minproduct * A [I], 1); minproduct = temp * A [I];} r = max (R, maxproduct);} return r ;} int main (INT argc, char * argv []) {int A [] = {1,-2,-1, 0, 5}; int result = maxproduct (A, 5 ); cout <result <Endl; System ("pause"); Return 0 ;}

2.14 obtain the largest sum of sub-Arrays
Given an array, all elements are integers (positive and negative), and they are used to find continuous elements and sum them into the largest sequence.
Http://blog.csdn.net/aoxiangzhiguanjun/article/details/8836702

3.2 telephone numbers correspond to English words and can be queried from the digital dictionary

# Include <iostream> using namespace STD; # define Tellen 3 void match (char * words) {char * word = "Yes yer"; if (strstr (word, words )) {printf ("words % s \ n", words) ;}} void main () {char word [Tellen + 1] = {0 }; // store each generated word char C [10] [10] = {"", "", "ABC", "def", "Ghi", "jkl ", "MnO", "pqrs", "TUV", "wxyz"}; // int total [10] = {, 3, 3, 3, 3, 4, 3, 4}; // number of letters in each number int number [Tellen] = {9, 3, 7}; // phone number int answer [10] = {0 }; // The Array records the offset (INDEX) of each letter in the character set that can be represented by its numeric key, initialized to 0int I, j = 0; while (true) {for (I = 0; I <Tellen; I ++) {If (number [I] = 1 | Number [I] = 0) // ignore the influence of spaces on break; else {printf ("% C", C [number [I] [answer [I]); word [I] = C [number [I] [answer [I] ;}} word [Tellen] = '\ 0'; match (Word ); // match printf ("\ n") in the data dictionary; int K = telLen-1; while (k> = 0) {If (answer [k] <total [number [k]-1) {answer [k] ++; break;} else {answer [k] = 0; k -- ;}} if (k <0) break;} system ("pause ");}

3.6 programming to determine whether two linked lists are intersecting
Http://blog.csdn.net/aoxiangzhiguanjun/article/details/8804403

3.8 Binary Tree Problems
Http://blog.csdn.net/aoxiangzhiguanjun/article/details/8904794

3.9 rebuilding a binary tree

# Include <stdio. h> # include <stdlib. h> # include <string. h> typedef struct node {char chvalue; struct node * lchild; struct node * rchild;} node; // rebuilt Binary Tree void rebuild (char * ppreorder, char * pinorder, node ** proot, int ntreelen) {int nleftlen, nrightlen; char * pleftend; node * P; // check if (! Ppreorder |! Pinorder |! Proot) return; If (! (P = (node *) malloc (sizeof (node) return; P-> chvalue = * ppreorder; P-> lchild = p-> rchild = NULL; * proot = P; If (ntreelen = 1) return; // divide the Left and Right subquantities pleftend = pinorder; while (* pleftend! = * Ppreorder) pleftend ++; nleftlen = (INT) (pleftend-pinorder); nrightlen = ntreelen-nleftlen-1; if (nleftlen) rebuild (ppreorder + 1, pinorder, & (p-> lchild), nleftlen); If (nrightlen) rebuild (ppreorder + nleftlen + 1, pinorder + nleftlen + 1, & (p-> rchild ), nrightlen);} // post-order traversal void postorder (node * P) {If (p) {postorder (p-> lchild); postorder (p-> rchild ); printf ("% C", p-> chvalue) ;}} int main (void) {char preo Rder [32], inorder [32]; node * ptree; // enter the first and middle order sequences while (scanf ("% S % s", preorder, inorder )! = EOF) // abdcef dbaecf {rebuild (preorder, inorder, & ptree, strlen (preorder); postorder (ptree); printf ("\ n ");} return 0 ;}


4.9 Data independence Construction

# Include <iostream> # include <cstdlib> using namespace STD;/* problem: To construct a 9*9 square matrix, players must be in each square, fill in any number ranging from 1 to 9 so that the numbers in each column, row, and 3x3 matrix of the entire board are not repeated. First, we use a deep priority search to generate a feasible solution, and then randomly delete a certain number of numbers to generate a Sudoku. */# Define Len 9 # define clear (a) memset (a), 0, sizeof (A) int level [] = {30, 37, 45 }; int grid [Len + 1] [Len + 1]; int value [Len + 1]; void next (Int & X, Int & Y) {x ++; if (x> 9) {x = 1; y ++ ;}// select the next valid state int picknextvalidvalue (int x, int y, int cur) {clear (value); int I, j; for (I = 1; I <Y; I ++) value [grid [I] [x] = 1; for (j = 1; j <X; j ++) value [grid [y] [J] = 1; int u = (x-1)/3*3 + 1; int v = (Y-1)/3*3 + 1; for (I = V; I <v + 3; I ++) for (j = u; j <u + 3; j ++) {value [grid [I] [J] = 1 ;}for (I = cur + 1; I <= Len & value [I]; I ++); return I;} void pre (Int & X, Int & Y) {X --; if (x <1) {x = 9; y -- ;}} int times = 0; int main () {int X, Y, I, j; X = y = 1; // Iterative Algorithm for deep search while (true) {times ++; // If (y = Len & X = Len) {for (I = 1; I <= Len; I ++) {for (j = 1; j <= Len; j ++) cout <grid [I] [J] <"; cout <Endl;} cout <Times <Endl; break; // pre (x, y ); // times = 0;} // If (y = 0) break if the failure result is met; // change the status grid [y] [x] = picknextvalidvalue (x, y, grid [y] [x]); If (grid [y] [x]> Len) {// restore status grid [y] [x] = 0; Pre (X, y);} else // further search for next (x, y);} for (I = 1; I <= level [2]; I ++) {int ind = rand () % (LEN * Len); grid [ind/Len + 1] [ind % Len] = 0 ;}for (I = 1; I <= Len; I ++) {for (j = 1; j <= Len; j ++) cout <grid [I] [J] <""; cout <Endl;} system ("pause ");}


4.10 digital puzzles and replies
# Include <iostream> # include <string> using namespace STD; // Title: Great Buddha Temple * I = great Buddha Temple. each letter represents a different number. int main () {bool flag; bool isused [10]; int number, revert_number, T, V; For (number = 0; number <100000; number ++) {flag = true; memset (isused, 0, sizeof (isused); t = number; revert_number = 0; For (INT I = 0; I <5; I ++) {v = T % 10; revert_number = revert_number * 10 + V; T/= 10; If (isused [v]) Flag = false; elseisused [v] = 1 ;} if (flag & (revert_number % Number = = 0) {v = revert_number/number; If (v <10 &&! Isused [v]) cout <number <"" <v <"" <revert_number <Endl ;}} system ("pause"); Return 0 ;}

The beauty of Programming
For some of the above Code, refer to the beauty of programming, and refer to the network, while others are compiled by yourself.