[置顶] 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 3
int 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 6
void 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];
}
else
break;
}
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";
else
cout<<"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 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};//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0
int 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");
}
#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;
else
IsUsed[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。