算法及编程语言 - 声振论坛 - 振动,动力学,声学,信号处理,故障诊断

chidan 发表于 2006-3-7 15:30

平方根.c

#define Epsilon 1.0E-6/*控制解的精度*/ #include <stdio.h> #include <math.h> 
main() { float num,pre,this; do { scanf("%f",&num);/*输入要求平方根的数*/ }while(num<0); if (num==0) printf("the root is 0"); else { this=1; do { pre=this; this=(pre+num/pre)/2; }while(fabs(pre-this)>Epsilon);/*用解的精度，控制循环次数，fabs()是求绝对值的函数*/ } printf("the root is %f",this); }

chidan 发表于 2006-3-7 15:32

圆周率.c

#include<stdio.h> long a=10000,b,c=2800,d,e,f[ 2801 ],g; main() { for(;b-c;) f[ b++ ]=a/5; for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a) for(b=c;d+=f[ b ]*a,f[ b ]=d%--g,d/=g--,--b;d*=b); }

chidan 发表于 2006-3-7 15:33

灯塔问题.c

//灯塔问题 #include <iostream.h> #include <fstream.h> #include <conio.h> int sz,cf=1,k,n,a,b,c; void shuru(void); void shuchu(void); bool panduan(void); void goujian(void); void main() { int i,j,lj=0,d; shuru(); for(i=1;i<=n;i++)cf=cf*2; for(i=0;i<cf;i++) { d=i; for(j=1;j<=n;j++) {sz=d%2;d=d/2;} goujian(); if(panduan()==true){lj=lj+1;shuchu();} } cout<<"共有"<<lj<<"种情况"<<endl; getch(); } void goujian(void) { int i1,j1; for(i1=n-1;i1>0;i1--) { for(j1=1;j1<=i1;j1++) { if(sz==1&&sz==1) sz=0; if(sz==0&&sz==0) sz=0; if(sz==1&&sz==0) sz=1; if(sz==0&&sz==1) sz=1; } } } bool panduan() { int pd=1,j1; for(j1=1;j1<=k;j1++) if(sz]]!=c) pd=0; if(pd==0) return false;else return true; } void shuchu(void) { int i2,j2; for(i2=1;i2<=n;i2++) { for(j2=1;j2<=n-i2;j2++) cout<<" "; for(j2=1;j2<=i2;j2++) cout<<sz<<" "; cout<<endl; } cout<<endl; } void shuru(void) { // char filename; ifstream input; // cout<<"Input filename:"; // cin>>filename; // input.open(filename); input.open("dt.txt"); input>>n; k=0; do{ k++; input>>a>>b>>c; }while((a!=0)&&(b!=0)); k--; }

chidan 发表于 2006-3-7 15:34

16进制10进制.c

//返回16进制字符串s对应的整数值，遇到任何一个非法字符都返回-1。 int HexToDec(char *s) { char *p = s;
 //空串返回0。 if(*p == '\0') return 0; //忽略开头的'0'字符 while(*p == '0') p++;
 int dec = 0; char c;
 //循环直到字符串结束。 while(c = *p++) { //dec乘16 dec <<= 4; //数字字符。 if(c >= '0' && c <= '9') { dec += c - '0'; continue; }
//小写abcdef。 if(c >= 'a' && c <= 'f') { dec += c - 'a' + 10; continue; }
//大写ABCDEF。 if(c >= 'A' && c <= 'F') { dec += c - 'A' + 10; continue; }
//没有从任何一个if语句中结束，说明遇到了非法字符。 return -1; }
 //正常结束循环，返回10进制整数值。 return dec; }

chidan 发表于 2006-3-7 15:35

矩阵转换.c

void trans(int *p,int n) { int i,j,temp; int *pi,*pj; for(i=0;i<=n-1;i++) { for(j=0;j<=i;j++) { pi =p +i*n;/*p首地址 */ pj =p +j*n; temp=pi; pi=pj; pj=temp; }
 } return; }
main() { int a={ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12, 13,14,15,16}; int i,j; printf("before transform:\n"); for(i=0;i<=3;i++) { for(j=0;j<=3;j++) printf("%d",a); printf("\n"); } trans((int*)a,4); printf("after transform:\n"); for(i=0;i<=3;i++) { for(j=0;j<=3;j++) printf("%d",a); printf("\n"); } return; }

chidan 发表于 2006-3-7 15:35

杨辉三角形.c

#include <stdio.h> int c(x,y); main() { int i,j,n=13; printf("N="); while(n>12) scanf("%d",&n); for(i=0;i<=n;i++) { for(j=0;j<12-i;j++) printf(" "); for(j=1;j<i+2;j++) printf("%6d",c(i,j)); printf("\n"); } }
int c(x,y) int x,y; { int z; if((y==1)||(y==x+1))return(1); z=c(x-1,y-1)+c(x-1,y);
return(z); }

chidan 发表于 2006-3-7 15:36

数组操作.c

#include <stdio.h>
void main() { char strg,*there,one,two; int *pt,list,index;
 strcpy(strg,"This is a character string.");
 one = strg; /* one 和 two是相同的 */ two = *strg; printf("第一输出的是 %c %c\n",one,two);
 one = strg; two = *(strg+8); printf("第二输出的是 %c %c\n",one,two);
 there = strg+10; /* strg+10 等同于 strg */ printf("第三输出的是 %c\n",strg); printf("第四输出的是 %c\n",*there);
 for (index = 0;index < 100;index++) list = index + 100; pt = list + 27; printf("第五输出的是 %d\n",list); printf("第六输出的是 %d\n",*pt); }

chidan 发表于 2006-3-7 15:36

桶排序.c

#include<stdio.h>
void comp(int k[],int m,int l) { int i=10,j=0,z=1,y=1,x,w,b; for(w=0;w<m;w++) for(x=0;x<10;x++) { b=-1; } while(z>0) { z=l/i; i=i*10; ++j;//记录最大数的位数 } i=10; while(j>0) { for(z=0;z<=m;z++) { x=(k/y)%i; b=k; } w=0; for(z=0;z<10;z++) for(x=0;x<m;x++) { if(b>=0) { k=b; b=-1; w++; } } --j; y=y*10; } for(z=0;z<m;z++) { printf("%d ",k); } }
main() { int n,m=0,l=0; int a; printf("请输入正整数,负数表示结束:"); scanf("%d",&n); while(n>=0) { a=n; ++m; if(n>l) l=n;//记录最大数 scanf("%d",&n); }
 comp(a,m,l); }

chidan 发表于 2006-3-7 15:37

读写文本文件.c

#include "stdio.h" void main() { FILE *funny,*printer,*fp; char c;
 funny = fopen("TENLINES.TXT","r"); /*打开文本文件 */ printer = fopen("PRN","w"); /*开启打印机*/ fp = fopen("weew.TXT","w"); do{ c = getc(funny); /* 从文件中得到一个字符 */ if(c != EOF) { putchar(c); //屏幕上显示字符 putc(c,printer); // 打印机上打印字符 } }while (c != EOF); /*重复直到 EOF (end of file)*/ fprintf(fp,"%s",*funny); fclose(funny); fclose(printer); fclose(fp);//关闭打印机 }

chidan 发表于 2006-3-7 15:38

文件复制.c

#include<stdio.h> #include<stdlib.h>
void main(int argc,char *argv[]) { FILE *in,*out;
 if(argc!=3){ printf("\n Usage: Hcopy sourcefile targetfile.\n"); exit(1); } if((in=fopen(argv,"rb"))==NULL){ printf("\n Cannot open the source file.\n"); exit(2); } if((out=fopen(argv,"wb"))==NULL){ printf("\n Cannot open the targetfile.\n"); exit(3); }
 /* start copy */ while(!feof(in)) putc(getc(in),out);
 fclose(in); fclose(out); }

chidan 发表于 2006-3-7 15:38

文件加密.c

给文件加密的技术很多,其中又分为不同等级,以适合不同场合的需要.这里给出最简单的文件加密技术,即采用文件逐字节与密码异或方式对文件进行加密,当解密时,只需再运行一遍加密程序即可.
下面是一个实例程序,能对任意一个文件进行加密,密码要求用户输入,限8位以内(当然你可以再更改).程序有很好的容错设计,这是我们应该学习的.
/* Turbo 2.0 pass. give file a password! */
#include<stdio.h> #include<stdlib.h> #include<conio.h> #include<string.h>
void dofile(char *in_fname,char *pwd,char *out_fname);/*对文件进行加密的具体函数*/
void main(int argc,char *argv[])/*定义main()函数的命令行参数*/ { char in_fname;/*用户输入的要加密的文件名*/ char out_fname; char pwd;/*用来保存密码*/
 if(argc!=4){/*容错处理*/ printf("\nIn-fname:\n"); gets(in_fname);/*得到要加密的文件名*/
 printf("Password:\n"); gets(pwd);/*得到密码*/
 printf("Out-file:\n"); gets(out_fname);/*得到加密后你要的文件名*/ 
dofile(in_fname,pwd,out_fname); } else{/*如果命令行参数正确,便直接运行程序*/ strcpy(in_fname,argv); strcpy(pwd,argv); strcpy(out_fname,argv); dofile(in_fname,pwd,out_fname); }
}
 /*加密子函数开始*/ void dofile(char *in_fname,char *pwd,char *out_file) { FILE *fp1,*fp2; register char ch; int j=0; int j0=0;
 fp1=fopen(in_fname,"rb"); if(fp1==NULL){ printf("cannot open in-file.\n"); exit(1);/*如果不能打开要加密的文件,便退出程序*/ } fp2=fopen(out_file,"wb"); if(fp2==NULL){ printf("cannot open or create out-file.\n"); exit(1);/*如果不能建立加密后的文件,便退出*/ } while(pwd[++j0]); ch=fgetc(fp1);
/*加密算法开始*/ while(!feof(fp1)){ fputc(ch^pwd,fp2);/*异或后写入fp2文件*/ ch=fgetc(fp1); } fclose(fp1);/*关闭源文件*/ fclose(fp2);/*关闭目标文件*/ }
/*程序结束*/

chidan 发表于 2006-3-7 15:39

文件连接.c

/****************fcat.c***************/
#include<stdio.h> #include<stdlib.h> #define BUFSIZE 256
void main(int argc,char *argv[]) { int i; char buff; FILE *fp1,*fp2;
 if(argc==1){ printf("Rsage: fcat filename linke_fname."); printf("\n"); exit(1); } if((fp1=fopen (argv,"a"))==NULL){ printf("file %s cannot opened.\n",argv); exit(1); } for(i=2;i<argc;i++){ if((fp2=fopen(argv,"r"))==NULL){ printf("file %s cannot opened.\n"); exit(1); } while(fgets(buff,BUFSIZE,fp2)!=NULL) fputs(buff,fp1); fclose(fp2); } fclose(fp1); }

chidan 发表于 2006-3-7 15:40

网络最短路径Dijkstra算法.c

/* * File: shortest.c * Description: 网络中两点最短路径 Dijkstra 算法 * Shortest Path Dijkstra Algorithm * Created: 2001/11/25 * Author: Justin Hou */
#include <stdio.h> #define true1 #define false 0 #define I9999 /* 无穷大 */ #define N20 /* 城市顶点的数目 */
int cost = { {0,3,I,I,I,1,I,I,I,I,I,I,I,I,I,I,I,I,I,I}, {3,0,5,I,I,I,6,I,I,I,I,I,I,I,I,I,I,I,I,I}, {I,5,0,4,I,I,I,1,I,I,I,I,I,I,I,I,I,I,I,I}, {I,I,4,0,2,I,I,I,6,I,I,I,I,I,I,I,I,I,I,I}, {I,I,I,2,0,I,I,I,I,7,I,I,I,I,I,I,I,I,I,I}, {1,I,I,I,I,0,1,I,I,I,2,I,I,I,I,I,I,I,I,I}, {I,6,I,I,I,1,0,6,I,I,I,7,I,I,I,I,I,I,I,I}, {I,I,1,I,I,I,6,0,2,I,I,I,3,I,I,I,I,I,I,I}, {I,I,I,6,I,I,I,2,0,8,I,I,I,4,I,I,I,I,I,I}, {I,I,I,I,7,I,I,I,8,0,I,I,I,I,5,I,I,I,I,I}, {I,I,I,I,I,2,I,I,I,I,0,4,I,I,I,3,I,I,I,I}, {I,I,I,I,I,I,7,I,I,I,4,0,3,I,I,I,4,I,I,I}, {I,I,I,I,I,I,I,3,I,I,I,3,0,3,I,I,I,5,I,I}, {I,I,I,I,I,I,I,I,4,I,I,I,3,0,7,I,I,I,2,I}, {I,I,I,I,I,I,I,I,I,5,I,I,I,7,0,I,I,I,I,3}, {I,I,I,I,I,I,I,I,I,I,3,I,I,I,I,0,5,I,I,I}, {I,I,I,I,I,I,I,I,I,I,I,4,I,I,I,5,0,8,I,I}, {I,I,I,I,I,I,I,I,I,I,I,I,5,I,I,I,8,0,6,I}, {I,I,I,I,I,I,I,I,I,I,I,I,I,2,I,I,I,6,0,4}, {I,I,I,I,I,I,I,I,I,I,I,I,I,I,3,I,I,I,4,0} }; int dist; /* 存储当前最短路径长度 */ int v0 = 'A' - 65; /* 初始点是 A */
void main() { int final, i, v, w, min;
 /* 初始化最短路径长度数据，所有数据都不是最终数据 */ for (v = 0; v < N; v++) { final = false; dist = cost; }
 /* 首先选v0到v0的距离一定最短，最终数据 */ final = true;
 /* 寻找另外 N-1 个结点 */ for (i = 0; i < N-1; i++) { min = I; /* 初始最短长度无穷大*/ /* 寻找最短的边 */ for (w = 0; w < N; w++) { if (!final && dist < min) { min = dist; v = w; } } final = true; /* 加入新边 */
 for (w = 0; w < N; w++) { /* 更新 dist[] 数据*/ if (!final && dist + cost < dist) { dist = dist + cost; } } }
 for (i = 0; i < N; i++) { /* 显示到监视器 */ printf("%c->%c: %2d\t", v0 + 65, i + 65, dist); } }

chidan 发表于 2006-3-7 15:40

矩阵乘法动态规划.c

/* * File: multi.c * Description:矩阵乘法动态规划 * Created: 10:20 2001-12-3 * Author: Justin Hou * */
#include <stdio.h> #defineN7
int middle;
void Show(int, int);
void main() { int i, j, l, k; unsigned long m, min; int r = {10, 20, 50, 1, 100, 4, 20, 2}; /* 矩阵维数 */
 /* 初始化 */ for (i = 1; i <= N; i++) { m = 0; } /* 每此增量加一 */ for (l = 1; l < N; l++) {
 /* 对于差值为 l 的两个元素 */ for (i = 1; i <= N - l; i++) { j = i + l; 
 /* 求其最小组合方式 */ min = m + m + r * r * r; middle = i; for (k = i + 1; k < j; k++) { if (min > m + m + r * r * r) { min = m + m + r * r * r; middle = k; } } m = min; } } printf("M = "); Show(1, N); printf("\nMultiply Count: %d\n", m); } void Show(int i, int j) { int k, m;
 if (i == j){ printf("M%d", i); /* 如果下一个是矩阵，输出*/ } else { m = middle; /* 分割成左右两组 */ if (i != m) printf("("); /* 如果下一个显示的不是矩阵 */ Show(i, m); /* 显示左边的内容 */ if (i != m) printf(")"); /* 如果上一个显示的不是矩阵 */ printf(" x "); /* 打印乘法符号 */ if (m+1 != j) printf("("); /* 如果下一个显示的不是矩阵 */ Show(m + 1, j); /* 显示右边的内容 */ if (m+1 != j) printf(")"); /* 如果下一个显示的不是矩阵 */ }
}

chidan 发表于 2006-3-7 15:41

万年历.c

#include "stdio.h" /* Required for MS-DOS use */ #define ENTER 0x1C0D/* Enter key */ int year, month, day; static char *days = {" ","Sunday ","Monday ","Tuesday", "Wednesday","Thursday ","Friday ","Saturday "}; struct TIMEDATE { int year; /* year 1980..2099 */ int month; /* month 1=Jan 2=Feb, etc. */ int day; /* day of month 0..31 */ int hours; /* hour 0..23 */ int minutes; /* minute 0..59 */ int seconds; /* second 0..59 */ int hsecs; /* 1/100ths of second 0..99 */ char dateline; /* date & time together */ }; static struct TIMEDATE today; main() { char cmonth; char cday; char cyear; double getdays(); double daynumb, numbnow; int weekday, retcode, dayer, i; dayer = datetime(&today); clrscn(); for (i=0;i<3;++i)cmonth='\0'; for (i=0;i<3;++i)cday='\0'; for (i=0;i<5;++i)cyear='\0'; putstr(5,8,14,"Enter date in MM DD YYYY format:"); while (retcode != ENTER) { retcode = bufinp(5,41,13,2,cmonth); if (retcode != ENTER) retcode = bufinp(5,44,13,2,cday); if (retcode != ENTER) retcode = bufinp(5,47,13,4,cyear); } year = atoi(&cyear); month = atoi(&cmonth); day = atoi(&cday); daynumb = getdays(year, month, day); numbnow = getdays(today.year, today.month, today.day); weekday = weekdays(daynumb); if (numbnow - daynumb == 0) printf("\n\n%02d-%02d-%d is",month, day, year); if (numbnow - daynumb > 0) printf("\n\n%02d-%02d-%d was",month, day, year); if (numbnow - daynumb < 0) printf("\n\n%02d-%02d-%d will be",month, day, year); printf(" a %s\n",days); }/* end MAIN */ /************************************************************ * GETDAYS- From integer values of year (YYYY), month * * (MM) and day (DD) this subroutine returns a * * double float number which represents the * * number of days since Jan 1, 1980 (day 1). * * This routine is the opposite of GETDATE. * ************************************************************/ double getdays(year, month, day) int year, month, day; { int y,m; double a,b,d, daynumb; double floor(),intg(); /********************************** ** make correction for no year 0 ** **********************************/ if (year < 0) y = year + 1; else y = year; /********************************************************* ** Jan and Feb are months 13 and 14 in this calculation ** *********************************************************/ m = month; if (month < 3) { m = m + 12; y = y - 1; } /************************** ** calculate Julian days ** **************************/ d = floor(365.25 * y) + intg(30.6001 * (m + 1)) + day - 723244.0; /********************************************** ** use Julian calendar if before Oct 5, 1582 ** **********************************************/ if (d < -145068.0) daynumb = d; /************************************* ** otherwise use Gregorian calendar ** *************************************/ else { a = floor(y / 100.0); b = 2 - a + floor(a / 4.0); daynumb = d + b; } return(daynumb); } /* end GETDAYS */ /******************************************************** * GETDATE - This routine takes a double float number * * representing the number of days since Jan 1,* * 1980 (day 1) and returns the year month and * * day as pointer integers * * This routine is the opposite of GETDAYS * ********************************************************/ getdate(numb) double numb; { double a,aa,b,c,d,e,z; double date; date = numb; z = intg(date + 2444239.0); if (date < -145078.0) a = z; else { aa = floor((z - 1867216.25) / 36524.25); a = z + 1 + aa - floor(aa/4.0); } b = a + 1524.0; c = intg((b - 122.1) / 365.25); d = intg(365.25 * c); e = intg((b - d) / 30.6001); day = b - d - intg(30.6001 * e); if (e > 13.5) month = e - 13.0; else month = e - 1.0; if (month > 2) year = c - 4716.0; else year = c - 4715.0; if (year < 1) --year; return; } /* end GETDATE */ /******************************************************** * WEEKDAYS - This routine takes a double float number* * representing the number of days since Jan 1,* * 1980 (day 1) and returns the day of the week* * where 1 = Sunday, 2 = Tuesday, etc. * ********************************************************/ int weekdays(numb) double numb; { double dd; int day; dd = numb; while (dd > 28000.0) dd = dd - 28000.0; while (dd < 0) dd = dd + 28000.0; day = dd; day = ((day + 1) % 7) + 1; return(day); } /******************************************************** * FRACT -This routine takes a double float number * * and returns the fractional part as a double * * float number * ********************************************************/ double fract(numb) double numb; { int inumb; double fnumb; while (numb < -32767) numb += 32767; while (numb > 32767) numb -= 32767; inumb = numb; fnumb = inumb; return(numb-fnumb); } /* end FRACT */ /******************************************************** * FLOOR -This routine takes a double float number * * and returns the next smallest integer * ********************************************************/ double floor(numb) double numb; { 
 double fract(), intg(); double out; out = intg(numb); if (numb < 0 && fract(numb) != 0) out -= 1.0; return(out); } /* end FLOOR */ /******************************************************** * INTG-This routine takes a double float number * * and returns the integer part as a double * * float number * ********************************************************/ double intg(numb) double numb; { double fract(); return(numb - fract(numb)); } /* end INTG */

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声振论坛's Archiver

平方根.c

圆周率.c

灯塔问题.c

16进制10进制.c

矩阵转换.c

杨辉三角形.c

数组操作.c

桶排序.c

读写文本文件.c

文件复制.c

文件加密.c

文件连接.c

网络最短路径Dijkstra算法.c

矩阵乘法动态规划.c

万年历.c