编程之美

1.1 cpu使用问题

#include <iostream>#include <ctime>#include <cmath>#include <Windows.h>using namespace std;//第一种方式void main(){INT64 start=0;int busy=10;int idle=busy;cout<<"CPU使用率问题";while(true){start=GetTickCount();while((GetTickCount()-start)<=busy);Sleep(idle);}}//第二种方式int main(){for(;;){for(int i = 0; i < 9600000; i++);//for(int i = 0; i < 21360000; i++);//2.67Ghz 4核Sleep(10);}return 0;}//正玄曲线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++;}}

CPU核心运行周期数

#include <iostream>using namespace std;inline __int64 GetCPUTickCount()  {  __asm  {  rdtsc;  }  }void main(){cout<<"CPU核心运行周期数"<<GetCPUTickCount()<<endl;system("pause");}

1.2 将帅问题

#include <iostream>using namespace std;//第一种方式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");}//第二种方式#define HALF_BITS_LENGTH 4// 这个值是记忆存储单元长度的一半，在这道题里是4bit#define FULLMASK 255// 这个数字表示一个全部bit的mask，在二进制表示中，它是11111111。#define LMASK (FULLMASK << HALF_BITS_LENGTH)// 这个宏表示左bits的mask，在二进制表示中，它是11110000。#define RMASK (FULLMASK >> HALF_BITS_LENGTH)// 这个数字表示右bits的mask，在二进制表示中，它表示00001111。#define RSET(b, n) (b = ((LMASK & b) ^ n))// 这个宏，将b的右边设置成n#define LSET(b, n) (b = ((RMASK & b) ^ (n << HALF_BITS_LENGTH)))// 这个宏，将b的左边设置成n#define RGET(b) (RMASK & b)// 这个宏得到b的右边的值#define LGET(b) ((LMASK & b) >> HALF_BITS_LENGTH)// 这个宏得到b的左边的值#define GRIDW 3// 这个数字表示将帅移动范围的行宽度。#include <stdio.h>#define HALF_BITS_LENGTH 4#define FULLMASK 255#define LMASK (FULLMASK << HALF_BITS_LENGTH)#define RMASK (FULLMASK >> HALF_BITS_LENGTH)#define RSET(b, n) (b = ((LMASK & b) ^ n))#define LSET(b, n) (b = ((RMASK & b) ^ (n << HALF_BITS_LENGTH)))#define RGET(b) (RMASK & b)#define LGET(b) ((LMASK & b) >> HALF_BITS_LENGTH)#define GRIDW 3int main(){unsigned char b;for(LSET(b, 1); LGET(b) <= GRIDW * GRIDW; LSET(b, (LGET(b) + 1)))for(RSET(b, 1); RGET(b) <= GRIDW * GRIDW; RSET(b, (RGET(b) + 1)))if(LGET(b) % GRIDW != RGET(b) % GRIDW)printf("A = %d, B = %d\n", LGET(b), RGET(b));system("pause");return 0;}

1.8 电梯调度

#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两堆石头

#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<<"输入两堆石头的书数目\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<<"先取石头玩家先赢\n";elsecout<<"后取石头玩家先赢\n";system("pause");}

 1.16 24点游戏

#include <iostream>#include <string>#include <cmath>#include <stdlib.h>using namespace std;const double PRECISION = 1E-6;const int COUNT_OF_NUMBER  = 4;const int NUMBER_TO_CAL = 24;double number[COUNT_OF_NUMBER];string expression[COUNT_OF_NUMBER];bool Search(int n){    if (n == 1) {    if ( fabs(number[0] - NUMBER_TO_CAL) < PRECISION ) {        cout << expression[0] << endl;        return true;    } else {        return false;    }}for (int i = 0; i < n; i++) {    for (int j = i + 1; j < n; j++) {        double a, b;        string expa, expb;        a = number[i];        b = number[j];        number[j] = number[n - 1];        expa = expression[i];        expb = expression[j];        expression[j] = expression[n - 1];        expression[i] = '(' + expa + '+' + expb + ')';        number[i] = a + b;        if ( Search(n - 1) ) return true;        expression[i] = '(' + expa + '-' + expb + ')';        number[i] = a - b;        if ( Search(n - 1) ) return true;        expression[i] = '(' + expb + '-' + expa + ')';        number[i] = b - a;        if ( Search(n - 1) ) return true;        expression[i] = '(' + expa + '*' + expb + ')';        number[i] = a * b;        if ( Search(n - 1) ) return true;        if (b != 0) {            expression[i] = '(' + expa + '/' + expb + ')';            number[i] = a / b;            if ( Search(n - 1) ) return true;        }        if (a != 0) {            expression[i] = '(' + expb + '/' + expa + ')';            number[i] = b / a;            if ( Search(n - 1) ) return true;        }        number[i] = a;        number[j] = b;        expression[i] = expa;        expression[j] = expb;    }}    return false;}int main(){    for (int i = 0; i < COUNT_OF_NUMBER; i++) {        char buffer[20];        int  x;        cin >> x;        number[i] = x;        itoa(x, buffer, 10);        expression[i] = buffer;    }    if ( Search(COUNT_OF_NUMBER) ) {        cout << "Success." << endl;    } else {        cout << "Fail." << endl;    }    return 0;}

 

2.2阶乘

void countZero(){    //100！末尾有多少个0    int num,i,count=0;    printf("Input a num as num!\n");    scanf("%d",&num);    for(i=5;i<=num;i+=5)    {        count++;        if(!(num%25))            count++;    }    printf("end of %d! has %d zero\n",num,count);    count=0;    while(num)    {        num/=2;        count+=num;    }    printf("n! last one %d\n",count+1);//n!最低位1的位置 等同于求n！中有多少个质因数2}int main(){    countZero();//计算100！末尾0的个数     return 0;}

 

2.4 1的数目

int count1Int(int i){    int numi=0;    while(i!=0)    {        numi+=(i%10==1)?1:0;        i/=10;    }    return numi;}void countOne(){    int i,n,count=0;    printf("please input a number\n");    scanf("%d",&n);    for(i=1;i<=n;i++)        count+=count1Int(i);    printf("count one %d",count);}int main(){    countOne();//一的数目    return 0;}

 

2.7 最大公约数最小公倍数求解

 

2.13 子数组最大乘积

#include <iostream>  #include   <stdlib.h>   #include   <stdio.h>   using namespace  std;  // 子数组的最大乘积  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 求子数组最大和

给一个数组，元素都是整数（有正数也有负数），寻找连续的元素相加之和为最大的序列。

 

3.2 电话号码对应英语单词并实现从数字字典中查询

#include <iostream>using namespace std;#define telLen 3void match(char *words){char *word="YES YER";if(strstr(word,words)){printf("words %s\n",words);}}void main(){char word[telLen+1]={0};//存储生成的每个单词char c[10][10]={"","","ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};//0到9数字所表示的字母int total[10]={0,0,3,3,3,3,3,4,3,4};//每个数字里包含的字母个数int number[telLen]={9,3,7};//电话号码int answer[10]={0};//数组记录每个字母在它所在的数字键能代表的字符集中的偏移（索引）,初始化为0int i,j=0;while(true){for (i=0;i<telLen;i++){if(number[i]==1||number[i]==0)//忽略空格的影响break;else{printf("%c",c[number[i]][answer[i]]);word[i]=c[number[i]][answer[i]];}}word[telLen]='\0';match(word);//在数据字典中匹配printf("\n");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 编程判断两个链表是否相交

3.8 二叉树中的一些问题

3.9 重建二叉树

#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct Node{char    chValue;struct Node    *lChild;struct Node    *rChild;}Node;//重建二叉树void Rebuild(char *pPreOrder , char *pInOrder , Node **pRoot , int nTreeLen){int  nLeftLen , nRightLen;char *pLeftEnd;Node *p;//边界条件检查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;//划分左右子数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);}//后序遍历void PostOrder(Node *p){if(p){PostOrder(p->lChild);PostOrder(p->rChild);printf("%c",p->chValue);}}int main(void){char PreOrder[32] , InOrder[32];Node *pTree;//输入先序和中序序列while(scanf("%s%s", PreOrder , InOrder) != EOF)//abdcef   dbaecf{Rebuild(PreOrder , InOrder , &pTree , strlen(PreOrder));PostOrder(pTree);printf("\n");}return 0;}

4.9 数独的构造

#include <iostream>#include <cstdlib>#include <cstring>using namespace std;/*问题：构造一个9*9的方格矩阵，玩家要在每个方格中，分别填上1至9的任意一个数字，让整个棋盘每一列、每一行以及每一个3*3的小矩阵中的数字都不重复。首先我们通过一个深度优先搜索来生成一个可行解，然后随机删除一定数量的数字，以生成一个数独。*/#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++;    }}// 选择下一个有效状态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;    // 深度搜索的迭代算法    while (true)    {        times++;        // 满足成功结果        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;        }        // 满足失败结果        if (y==0)            break;        // 改变状态        grid[y][x] = pickNextValidValue(x, y, grid[y][x]);        if (grid[y][x] > LEN)        {            // 恢复状态            grid[y][x] = 0;            pre(x, y);        }        else            // 进一步搜索            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 数字哑谜和回文

#include<iostream>#include<string>using namespace std;// 题目:人过大佛寺*我=寺佛大过人.   其中每个字母代表着一个不同的数字.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;}

编程之美 完

以上有些代码参考《编程之美》还有一些是参考网络上的，剩下的是自己编写的。