各种基本算法实现小结（七）—— 常用算法

 

转载地址：http://blog.csdn.net/sunboy_2050/article/details/5645837

 

各种基本算法实现小结（七）—— 常用算法

（均已测试通过）

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1、判断素数

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <math.h> 
3. int is_sushu(int n) 
4. { 
5.     int i, mid; 
6.     mid=(int)sqrt(n); 
7.     for(i=2; i<=mid; i++) 
8.         if(0 == n%i) 
9.             return 0; 
10.     return 1; 
11. } 
12. void main() 
13. { 
14.     int n; 
15.      
16.     printf("Enter a num: "); 
17.     scanf("%d", &n); 
18.     if(is_sushu(n)) 
19.         printf("%d is sushu!/n", n); 
20.     else 
21.         printf("%d is not sushu.../n", n); 
22. } 

#include <stdio.h>#include <math.h>int is_sushu(int n){int i, mid;mid=(int)sqrt(n);for(i=2; i<=mid; i++)if(0 == n%i)return 0;return 1;}void main(){int n;printf("Enter a num: ");scanf("%d", &n);if(is_sushu(n))printf("%d is sushu!/n", n);elseprintf("%d is not sushu.../n", n);}

运行结果：

   

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2、 求2-1000之间的所有素数

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <math.h> 
3. #define MAX 1000 
4. int is_sushu(int n) 
5. { 
6.     int i, mid; 
7.     mid=(int)sqrt(n); 
8.     for(i=2; i<=mid; i++) 
9.         if(0 == n%i) 
10.             return 0; 
11.     return 1; 
12. } 
13. void main() 
14. { 
15.     int i, count; 
16.      
17.     count=0; 
18.      
19.     for(i=2; i<=MAX; i++) 
20.         if(is_sushu(i)) 
21.         { 
22.             count++; 
23.             printf("%5d", i); 
24.                 if(0 == count%10) 
25.                     printf("/n"); 
26.         } 
27.     printf("/n"); 
28. } 

#include <stdio.h>#include <math.h>#define MAX 1000int is_sushu(int n){int i, mid;mid=(int)sqrt(n);for(i=2; i<=mid; i++)if(0 == n%i)return 0;return 1;}void main(){int i, count;count=0;for(i=2; i<=MAX; i++)if(is_sushu(i)){count++;printf("%5d", i);if(0 == count%10)printf("/n");}printf("/n");}

运行结果：

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3、 验证哥德巴赫猜想

哥德巴赫猜想：任意一个大于等于6的偶数都可以分解为两个素数之和

如： 6 = 3+3；100 = 3+97=11+89； 1000 = 3+997=59+941=。。。

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <math.h> 
3. #define MAX 1000 
4. int is_sushu(int n) 
5. { 
6.     int i, mid; 
7.     mid=(int)sqrt(n); 
8.     for(i=2; i<=mid; i++) 
9.         if(0 == n%i) 
10.             return 0; 
11.     return 1; 
12. } 
13. void main() 
14. { 
15.     int i, mid, n; 
16.      
17.     printf("Enter an even num: "); 
18.     scanf("%d", &n); 
19.      
20.     mid=n/2; 
21.     for(i=2; i<=mid; i++) 
22.     { 
23.         if(is_sushu(i) && is_sushu(n-i)) 
24.             printf("%d = %d + %d/n", n, i, n-i); 
25.     } 
26.      
27. } 

#include <stdio.h>#include <math.h>#define MAX 1000int is_sushu(int n){int i, mid;mid=(int)sqrt(n);for(i=2; i<=mid; i++)if(0 == n%i)return 0;return 1;}void main(){int i, mid, n;printf("Enter an even num: ");scanf("%d", &n);mid=n/2;for(i=2; i<=mid; i++){if(is_sushu(i) && is_sushu(n-i))printf("%d = %d + %d/n", n, i, n-i);}}

运行结果：

     

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4、 求最大公约数（GCD）和最小公倍数(LCM)

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. void  max_min(int &m,int &n) 
3. { 
4.     int tmp; 
5.     if(m<n) 
6.     { 
7.         tmp=m; 
8.         m=n; 
9.         n=tmp; 
10.     } 
11. } 
12. int Cal_GCD(int m,int n) 
13. { 
14.     int gcd; 
15.      
16.     max_min(m, n); 
17.     gcd=m%n; 
18.     while(gcd) 
19.     { 
20.         m=n; 
21.         n=gcd; 
22.         gcd=m%n; 
23.     } 
24.     return n; 
25. } 
26. void main() 
27. { 
28.     int m, n, gcd; 
29.      
30.     printf("Enter two num a b: "); 
31.     scanf("%d %d", &m, &n); 
32.     gcd=Cal_GCD(m, n); 
33.     printf("%d and %d GCD: %d/n", m, n, gcd); 
34.     printf("%d and %d LCM: %d/n", m, n, m*n/gcd); 
35. } 

#include <stdio.h>void max_min(int &m, int &n){int tmp;if(m<n){tmp=m;m=n;n=tmp;}}int Cal_GCD(int m, int n){int gcd;max_min(m, n);gcd=m%n;while(gcd){m=n;n=gcd;gcd=m%n;}return n;}void main(){int m, n, gcd;printf("Enter two num a b: ");scanf("%d %d", &m, &n);gcd=Cal_GCD(m, n);printf("%d and %d GCD: %d/n", m, n, gcd);printf("%d and %d LCM: %d/n", m, n, m*n/gcd);}

运行结果：

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5、统计个数（数字）

用随机函数产生100个[0，99]范围内的随机整数，

统计个位上的数字分别为0，1，2，3，4，5，6，7，8，9的数的个数并打印出来

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <stdlib.h> 
3. #include <time.h> 
4. #include <string.h> 
5. #define MAX 101 
6. void input(int num[]) 
7. { 
8.     int i; 
9.     srand((unsigned)time(NULL)); 
10.     for(i=1; i<MAX; i++) 
11.         num[i]=rand()%100; 
12. } 
13. void output(int num[]) 
14. { 
15.     int i; 
16.     for(i=1; i<MAX; i++) 
17.     { 
18.         printf("%5d", num[i]); 
19.         if(0==i%10) 
20.             printf("/n"); 
21.     } 
22.     printf("/n"); 
23. } 
24. void cal_num(int num[],int count[]) 
25. { 
26.     int i, mod; 
27.      
28.     for(i=1; i<MAX; i++) 
29.     { 
30.         mod=num[i]%10; 
31.         count[mod]++; 
32.     } 
33. } 
34. void main() 
35. { 
36.     int num[MAX]; 
37.     int i, count[10]; 
38.      
39.     memset(count, 0, 10*sizeof(int));/* initial count[] to 0 */ 
40.     input(num); 
41.     printf("100 num:/n"); 
42.     output(num); 
43.      
44.     cal_num(num, count); 
45.     for(i=0; i<10; i++) 
46.         printf("%d: %d/n", i, count[i]);     
47. } 

#include <stdio.h>#include <stdlib.h>#include <time.h>#include <string.h>#define MAX 101void input(int num[]){int i;srand((unsigned)time(NULL));for(i=1; i<MAX; i++)num[i]=rand()%100;}void output(int num[]){int i;for(i=1; i<MAX; i++){printf("%5d", num[i]);if(0==i%10)printf("/n");}printf("/n");}void cal_num(int num[], int count[]){int i, mod;for(i=1; i<MAX; i++){mod=num[i]%10;count[mod]++;}}void main(){int num[MAX];int i, count[10];memset(count, 0, 10*sizeof(int)); /* initial count[] to 0 */input(num);printf("100 num:/n");output(num);cal_num(num, count);for(i=0; i<10; i++)printf("%d: %d/n", i, count[i]);}

运行结果：

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6、统计个数（数字、字符、其它字符）

输入一行字符，统计其中有多少个数字、字符和其它字符

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <string.h> 
3. #define MAX 1024 
4. void cal_num(char *str,int count[]) 
5. { 
6.     char *pstr; 
7.     pstr=str; 
8.     while(*pstr) /* *pstr != 0 */ 
9.     { 
10.         if(*pstr>='0' && *pstr<='9') 
11.             count[0]++; 
12.         else if((*pstr>='a' && *pstr<='z') || (*pstr>='A' && *pstr<='Z')) 
13.             count[1]++; 
14.         else 
15.             count[2]++; 
16.          
17.         pstr++; 
18.     } 
19. } 
20. void main() 
21. { 
22.     char str[MAX]; 
23.     int i, count[3];  /* 0->num; 1->char; 2->others */ 
24.     memset(count, 0, 3*sizeof(int)); 
25.      
26.     printf("Enter a string: "); 
27.     scanf("%s", str); 
28.     cal_num(str, count); 
29.     for(i=0; i<3; i++) 
30.     { 
31.         switch(i) 
32.         { 
33.             case 0: 
34.                 printf("num: %d/n", count[i]); 
35.                 break; 
36.             case 1: 
37.                 printf("char: %d/n", count[i]); 
38.                 break; 
39.             case 2: 
40.                 printf("other: %d/n", count[i]); 
41.                 break; 
42.         } 
43.     } 
44. } 

#include <stdio.h>#include <string.h>#define MAX 1024void cal_num(char *str, int count[]){char *pstr;pstr=str;while(*pstr) /* *pstr != 0 */{if(*pstr>='0' && *pstr<='9')count[0]++;else if((*pstr>='a' && *pstr<='z') || (*pstr>='A' && *pstr<='Z'))count[1]++;elsecount[2]++;pstr++;}}void main(){char str[MAX];int i, count[3]; /* 0->num; 1->char; 2->others */memset(count, 0, 3*sizeof(int));printf("Enter a string: ");scanf("%s", str);cal_num(str, count);for(i=0; i<3; i++){switch(i){case 0:printf("num: %d/n", count[i]);break;case 1:printf("char: %d/n", count[i]);break;case 2:printf("other: %d/n", count[i]);break;}}}

运行结果：

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7、 数制转换（递归实现）

本算法仅实现了基数为2-16的数制转换

如果大家希望扩展范围，仅需要对基数表示字符case 进行扩展即可，如G、H、I ...

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. int flag=1; /* check: n/d == 0 */ 
3. void trans_num(int n,int d) 
4. { 
5.     int mod; 
6.     mod=n%d; 
7.     n=n/d; 
8.     while(flag && n) 
9.         trans_num(n,d); 
10.     flag=0; 
11.     switch(mod) 
12.     { 
13.         case 10: 
14.             printf("A"); 
15.             break; 
16.         case 11: 
17.             printf("B"); 
18.             break; 
19.         case 12: 
20.             printf("C"); 
21.             break; 
22.         case 13: 
23.             printf("D"); 
24.             break; 
25.         case 14: 
26.             printf("E"); 
27.             break; 
28.         case 15: 
29.             printf("F"); 
30.             break; 
31.         default: 
32.             printf("%d", mod);   
33.     } 
34.          
35. } 
36. void main() 
37. { 
38.     int n, d; 
39.     printf("Enter n d: "); 
40.     scanf("%d %d", &n, &d); 
41.      
42.     trans_num(n, d); 
43.     printf("/n");    
44. } 

#include <stdio.h>int flag=1; /* check: n/d == 0 */void trans_num(int n, int d){int mod;mod=n%d;n=n/d;while(flag && n)trans_num(n,d);flag=0;switch(mod){case 10:printf("A");break;case 11:printf("B");break;case 12:printf("C");break;case 13:printf("D");break;case 14:printf("E");break;case 15:printf("F");break;default:printf("%d", mod);}}void main(){int n, d;printf("Enter n d: ");scanf("%d %d", &n, &d);trans_num(n, d);printf("/n");}

运行结果：

       

算法改进

数制直接转为字符输出，扩展支持16进制以上的数制转换

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1. #include <stdio.h> 
2. int flag=1; /* check: n/d == 0 */ 
3. void trans_num(int n,int d) 
4. { 
5.     int mod; 
6.     mod=n%d; 
7.     n=n/d; 
8.     while(flag && n) 
9.         trans_num(n,d); 
10.     flag=0; 
11.     if(mod>=10) 
12.         mod=mod-10+65;  /* convert to char */ 
13.     else 
14.         mod=mod+48; 
15.     printf("%c", mod);  /* print char (%c) */    
16. } 
17. void main() 
18. { 
19.     int n, d; 
20.     printf("Enter n d: "); 
21.     scanf("%d %d", &n, &d); 
22.      
23.     trans_num(n, d); 
24.     printf("/n");    
25. } 

#include <stdio.h>int flag=1; /* check: n/d == 0 */void trans_num(int n, int d){int mod;mod=n%d;n=n/d;while(flag && n)trans_num(n,d);flag=0;if(mod>=10)mod=mod-10+65; /* convert to char */elsemod=mod+48;printf("%c", mod);/* print char (%c) */}void main(){int n, d;printf("Enter n d: ");scanf("%d %d", &n, &d);trans_num(n, d);printf("/n");}

运行结果（扩展进制）：

             

100 = 4*24+4            1000=1*24*24+17*24+16  10000=17*24*24+8*24+16        1000=27*36+28

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8、 数制转换（栈实现）

核心思想和递归实现类似，都是压栈的原理，实现较简单，请自己尝试实现

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9、 水仙花数

水仙花数简述： 水仙花数是指一个 n 位数 ( n≥3 )，它的每个位上的数字的 n 次幂之和等于它本身。

如：153=1^3+5^3+3^3（3位数）；1634=1^4+6^4+3^4+4^4（4位数）；54748=5^5+4^5+7^5+4^5+8^5（5位数）

判断任一3位数，是否为水仙花数

测试环境：GCC

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1. #include <stdio.h> 
2. main() 
3. { 
4.     int b, s, g, n, sum; 
5.     scanf("%d", &n); 
6.     b=n/100; 
7.     s=n/10%10; 
8.     g=n%10; 
9.     sum=b*b*b+s*s*s+g*g*g; 
10.     if(sum==n) 
11.         printf("Yes/n"); 
12.     else 
13.         printf("No/n"); 
14. } 

#include <stdio.h>main(){int b, s, g, n, sum;scanf("%d", &n);b=n/100;s=n/10%10;g=n%10;sum=b*b*b+s*s*s+g*g*g;if(sum==n)printf("Yes/n");elseprintf("No/n");}

运行结果（Redhat Linux）：

================================================

求4位数的水仙花数（1000<=X<=9999）

测试环境：VC 6.0 (C)

 

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1. #include <stdio.h> 
2. int main() 
3. { 
4.     int i,j,k,l,m,n; 
5.     for(i=1; i<=9; i++) 
6.         for(j=0; j<=9; j++) 
7.             for(k=0; k<=9; k++) 
8.                 for(l=0; l<=9; l++) 
9.                     if((i*1000+j*100+k*10+l)==i*i*i*i+j*j*j*j+k*k*k*k+l*l*l*l) 
10.                         printf("%d%d%d%d=%d^4+%d^4+%d^4*%d^4/n", i, j, k, l, i, j, k, l); 
11.     return 0; 
12. } 

#include <stdio.h>int main(){int i,j,k,l,m,n;for(i=1; i<=9; i++)for(j=0; j<=9; j++)for(k=0; k<=9; k++)for(l=0; l<=9; l++)if((i*1000+j*100+k*10+l)==i*i*i*i+j*j*j*j+k*k*k*k+l*l*l*l)printf("%d%d%d%d=%d^4+%d^4+%d^4*%d^4/n", i, j, k, l, i, j, k, l);return 0;}

运行结果：

================================================

思考：如果求得高精度大数的水仙花数，如8位、18位、28位的水仙花数（需考虑计算机精度，可采用数组或指针实现，大数计算）

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10、 大数计算

大数运算：参加的值和计算结果通常是以上百位数，上千位数以及更大长度之间的整数运算，早已超出了计算机能够表示数值的精度范围（2^32=4294967296或2^64=18446744073709551616）即64位机最大也才20位，因此需要想出其它的办法计算大数。

求任意两整数之和（1000位以内）

测试环境：VC 6.0 (C)

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1. #include <stdio.h> 
2. #include <string.h> 
3. #define MAX 1000 /* precision */ 
4. void input(char ch[]) 
5. { 
6.     scanf("%s", ch); 
7. } 
8. void add(char ch1[],char ch2[], char ch3[]) 
9. { 
10.     int len1, len2, len3, maxlen; 
11.     int sum, flag; 
12.     len1=strlen(ch1); 
13.     len2=strlen(ch2); 
14.     len3=maxlen=len1 >= len2 ? len1 : len2; 
15.     flag=0; /* jin wei */ 
16.      
17.     while(len1>=1 && len2>=1) 
18.     { 
19.         sum=ch1[len1-1]-'0' + ch2[len2-1]-'0' + flag;/* char -> int to calculate sum */ 
20.         flag=0; 
21.          
22.         if(sum>=10) 
23.         { 
24.             sum-=10; 
25.             flag=1; 
26.         } 
27.         ch3[maxlen-1]=sum + '0'; 
28.         len1--; 
29.         len2--; 
30.         maxlen--; 
31.     } 
32.     while(len1>=1) /* if num1[] is longer or maxer */ 
33.     { 
34.         sum=ch1[len1-1]-'0' + flag; 
35.         flag=0; 
36.          
37.         if(sum>=10) 
38.         { 
39.             sum-=10; 
40.             flag=1; 
41.         } 
42.         ch3[maxlen-1]=sum + '0'; 
43.         len1--; 
44.         maxlen--; 
45.     } 
46.      
47.     while(len2>=1) /* if num2[] is longer or maxer */ 
48.     { 
49.         sum=ch2[len2-1]-'0' + flag; 
50.         flag=0; 
51.          
52.         if(sum>=10) 
53.         { 
54.             sum-=10; 
55.             flag=1; 
56.         } 
57.         ch3[maxlen-1]=sum + '0'; 
58.         len2--; 
59.         maxlen--; 
60.     } 
61.     if(flag != 0) /* if flag, then print gaowei(jinwei) */ 
62.         printf("%d", flag); 
63.     for(int i=0; i<len3; i++)  
64.         printf("%c", ch3[i]);    
65.     printf("/n"); 
66. } 
67. int main() 
68. { 
69.     char ch1[MAX], ch2[MAX], ch3[MAX+1]; 
70.     memset(ch3, '0', sizeof(ch3)); 
71.     input(ch1); 
72.     input(ch2); 
73.     add(ch1, ch2, ch3); 
74.     return 0; 
75. } 

#include <stdio.h>#include <string.h>#define MAX 1000 /* precision */void input(char ch[]){scanf("%s", ch);}void add(char ch1[], char ch2[], char ch3[]){int len1, len2, len3, maxlen;int sum, flag;len1=strlen(ch1);len2=strlen(ch2);len3=maxlen=len1 >= len2 ? len1 : len2;flag=0; /* jin wei */while(len1>=1 && len2>=1){sum=ch1[len1-1]-'0' + ch2[len2-1]-'0' + flag; /* char -> int to calculate sum */flag=0;if(sum>=10){sum-=10;flag=1;}ch3[maxlen-1]=sum + '0';len1--;len2--;maxlen--;}while(len1>=1) /* if num1[] is longer or maxer */{sum=ch1[len1-1]-'0' + flag;flag=0;if(sum>=10){sum-=10;flag=1;}ch3[maxlen-1]=sum + '0';len1--;maxlen--;}while(len2>=1) /* if num2[] is longer or maxer */{sum=ch2[len2-1]-'0' + flag;flag=0;if(sum>=10){sum-=10;flag=1;}ch3[maxlen-1]=sum + '0';len2--;maxlen--;}if(flag != 0) /* if flag, then print gaowei(jinwei) */printf("%d", flag);for(int i=0; i<len3; i++) printf("%c", ch3[i]);printf("/n");}int main(){char ch1[MAX], ch2[MAX], ch3[MAX+1];memset(ch3, '0', sizeof(ch3));input(ch1);input(ch2);add(ch1, ch2, ch3);return 0;}

运行结果：

   

思考：请大家自己设计实现更复杂的大数减法、乘法、除法，求余、求幂、求最小公倍数等大数运算（提示：可用数组或链表）