第一章：求所有阶乘和数，大奖赛现场统分程序设计，三位阶乘和数，图形点扫描，递归求n的阶乘n!，求n的阶乘n!，舍罕王失算

//求所有阶乘和数
 long fac3 (int x) 
 {
    
    int i; 
    long p=1;
    for (i=1 ;i<=x;i++) 
        p*=i;
    return (p);
 }
 int main()
 {
    
 	int a, b, c, d, e, f, g;
    long m1, m2, m3, m4, m5, m6, n1, n2, n3, n4, n5, n6;
    printf(" 所有阶乘和数有： "); 
    for (a = 1; a <9;a++) 
    {
    
         if (a == fac3(a))
          printf("%d", a);
          for (b=0;b <= 9;b++) 
          {
     
              m1 = a * 10 + b;
              n1=fac3 (a) + fac3 (b);
             if(m1 == n1) 
             printf("%ld ", m1);
             for (c=0;c<=9; c++) 
             {
     
                 m2 = m1 * 10 + c;
                 n2 = n1 + fac3(c);
                 if(m2 == n2)
                 printf("%ld", m2); 
                 for (d=0; d <= 9; d++) 
                 {
     
                     m3 = m2 * 10 + d;
                     n3 = n2 +fa3c(d);
                     if (m3==n3) printf("%ld" , m3);
                     for(e=0; e <= 9; e++) 
                     {
    
                          m4 = m3 * 10 + e; 
                          n4 = n3 + fac3 (e);
                          if(m4==n4) 
                          printf("%ld", m4) ;
                          for(f = 0; f <= 9; f++) 
                          {
     
                               m5 = m4 * 10 + f;
                               n5 = n4 + fac3(f);
                               if( m5 == n5)
                               printf("%ld",m5);
                               for( g = 0; g <= 9; g++)
                               {
    
                                    m6 = m5 * 10 + g;
                                    n6 = n5 + fac3(g); 
                                    if( m6 == n6) 
                                     printf("%ld",m6);
                               }
                           }
                       }
                    }
                 } 
            }
    }
 	
 	return 0;
 }
   
 

结果：

 //大奖赛现场统分程序设计
 int main()
 {
    
 	int i, j, n, m, u, sh[40], ph[20];
	double a,b, c, max, min, uf, sf[40], pf[20],z[40], f[40][20];
	printf("请输入选手个数 (1<n<40): ");
	scanf ("%d", &n); 
	printf("请输入评委个数 (2<m<20)：");
	scanf ("%d", &m); 
	for (j=1; j <= m; j++)
	 ph[j] = j;
	for(i = 1; i <= n; i++) // 现场为选手评分，统分 
	{
    
	 	printf ("\n 第%d 个上场选手编号为： ",i);
		scanf("%d",&sh[i]);// 输入第i个上场选手编号 
		sf[i]=0; 
		max=0; 
		min=100; 
		for(j=1; j <= m; j++) 
		{
    
			printf(" 第%d个评委评分为：",j);
			scanf("%lf",&f[i][j]);
			sf[i] += f[i][j];
			 if (max < f[i][j]) 
			     max = f[i][j]; //输入第j个评委为第i个选手的评分 //统计最高分与最低分 
			if (min > f[i][j]) 
			    min = f[i][j];
		}
		 printf("\n 去掉一个最高分：%.3f",max);
		 printf("\n 去掉一个最低分：%.3f",min); 
		 sf[i]= (sf[i]-max-min) / (m-2); // 第i个选手的最后得分 
		 a = max-sf[i];
		 b = sf[i]-min; 
		 z[i] = (a>b) ?a:b; 
		 printf ("\n 编号为%d号选手",sh[i]);
		 printf(" 最后得分为：%.3f",sf[i]); // 宜布第 i个选手的得分 
    }
    
    
    
		 for (c=0, i=1; i <= n;i++) 
		   c += z[i] * z[i];
		   c = sqrt (c/n); //给评委评分 
		 for(j=1; j <= m; j++) 
		 {
    		 	
			pf[j]=0;
			 for (i=1; i <=n; i++) 
			    pf[j]+=(f[i][j]-sf[i]) * (f[i][j]-sf[i]);
			 pf[j] = 10.0 - sqrt(pf[j]/n) /c*5;
			  printf("\n 第%d 号评委得分为：%.3f",j,pf[j]);
		}
		for(i = 1; i <= n; i++)
		for(j = i +1; j <= n; j++)
		   if(sf[i] < sf[j])
		   {
    
		   	
		   	 uf = sf[i];
			 sf[i] = sf[j];
			 sf[j] = uf;
			 u = sh[i];
			 sh[i] = sh[j];
			 sh[j] = u;
		   }
 	
	   printf ("\n 参賽选手得分名次表");
	   printf("\n 选手编号 得分 名次");
	    for (i=1; i <= n; i++)
		 printf ("\n %d %.3f %d", sh[i], sf[i], i) ;
		 printf("\n");
		 for(j = 1;j <= m-1;j++) // 评委得分从高到低排序 数
		 for (i = j+1;i <= m;i++) 
		 if (pf[j] < pf[i]) 
		 {
    
		    uf = pf[j];
			pf[j] = pf[i];
			pf[i] = uf; 
			u = ph[j];
			ph[j] = ph[i];
			ph[i] = u; 
	  	 }
		printf("\n 评委得分名次表");
		printf ("\n 评委编号 得分 名次");
		 for(j=1;j <= m; j++) 
		    printf ("\n %d %.3f %d", ph[j], pf [j], j) ;
			 printf ("\n 竞赛现场统分结束，谢谢！\n");
 	
 	return 0;
  } 

结果：

  //三位阶乘和数
    long fac2 (int x) 
 {
    
    int i; 
    long p=1;
    for (i=1;i<=x;i++) 
    
    	p*=i;
    	
	
        
    return (p);
 }
   int main()
   {
    
   	
   	  int a, b, c, m,n;
     printf(" 三位阶乘和数有：");
     for (a = 1; a <= 6; a++) 
     for (b = 0; b <= 6; b++) 
     for (c=0; c <= 6; c++)
     {
    
         m = a * 100 + b * 10 + c;
         n = fac2(a) + fac2(b) + fac2(c);
         if (m==n) printf ("%d \n", m) ;
     }
   	
   	
	} 
   

结果：

//图形点扫描 求地图面积
 int main()
{
     
	FILE *fp;
	char fname[50]; //文件名长度在此约定不大于 50
	int i, j, x, y, k, s, w, t, a[12] [12] ;
	printf (" \n input file name: ");
	gets (fname);
	if ( (fp=fopen (fname, "r")) == NULL) 
	{
     
		printf( "The file was not opened n" ) ;
		return 0;
	}
	x = 11;
	y = 11;
	for (i=0; i <=y; ++i)
	{
    
	  printf("\n ");
   		for(j=0; j<=x;++j) 
		{
     
		 	fscanf(fp, "%d", &a[i][j]); //，从文件读数据到二维a数组
			printf("%d ",a[i][j]);
		}
	}
		 		
	for(i = 0; i <= x ;++i)
	{
    
		 if(a[0][i] == 0)
		   a[0][i] = 2;
		   if(a[y][i] == 0)
		   a[y][i] = 2; 
   }
   	for(i = 1; i <= y ;++i)
	{
    
		 if(a[i][0] == 0)
		   a[i][0] = 2;
		   if(a[i][x] == 0)
		   a[i][x] = 2; 
    }
	t = x * y;
		 
		 for (k=1;k<t; ++k) 
		 {
    
		 	w = 0;
		 	for (i = 1; i < y; ++i)
		 	for( j = 1; j < x; ++j)
		 	if((a[i][j-1] == 2 || a[i][j+1] == 2 || a[i-1][j] == 2 || a[i+ 1][j] == 2) && a[i][j] == 0)
		 	{
    
		 		a[i][j] = 2;
		 		w = 1;
		    }
	      }
	s = 0;
	printf("\n ");	
	for(i = 0; i <= y; ++i)
	{
    
			printf("\n ");
			for(j = 0; j <= x; ++j)
			{
    
				if(a[i][j] == 0)  s += 1;
				printf("%d", a[i][j]);
			}
		
	} 		
	printf("\n s = %d\n", s); 		
	fclose(fp);
	return 0;
}

结果：

文件路径：

文件内容：

PS：没有空格会乱码

//递归求n的阶乘n!
long fac1(int n)
{
    
	long f;
	if (n == 1) f = 1;
	else f = n * fac1(n - 1);
	return (f);
}
int main()
{
    
	int n;
	long y;
	printf(" 请输入n（n < 13）：");
	scanf("%d", &n);
	if (n > 12)
	{
    
		printf(" 输入的n超出上限12，请重新输入n。\n");
		return 0;
	}
	y = fac1(n);
	printf(" %d! = %ld\n", n, y);
	return 0;
}

结果：

//求n的阶乘n!
int main()
{
    
	int k, n;
	long t;
	printf(" 请输入n (n < 3)", &n);
	scanf("%d", &n);
	t = 1;
	for (k = 1; k <= n; k++)
		t = t * k;
	printf(" %d != %ld\n", n, t);

	return 0;
	
}

结果：

//舍罕王失算
int main()
{
    
    double t, s, v, p;
    int i, n;
    printf(" 请输入格数n：");
    scanf("%d", &n);
    t = 1;
    s = 1;
    for (i = 2; i <= n; i++)
    {
    
        t = t * 2;
        s += t;

    }
    v = s / 2.4e7;
    p = v / 2.48e9;
    if (n <= 40) printf(" 总麦粒数为：%.0f\n", s);
    else printf(" 总麦粒数为：%.3e\n", s);
    printf(" 小麦重量约为：%f吨\n", v);
    printf(" 约相当于世界粮食年总产量的%.0f倍\n", p);
    return 0;
	
}

结果：