2-1 位数

#include
     
      #include
      
       int main (){    int n;    for (n=100; n<=999; n++)    {        if(n == pow(n/100,3) + pow(n/10%10,3) + pow(n%10,3) )             printf("%d\n",n);           }        system("PAUSE");    return EXIT_SUCCESS;}
      
     

2-2 水仙花数

#include
     
      #include
      
       #include 
       
        int main (){    int n;    for (n=100; n<=999; n++)    {        if(n == pow(n/100,3) + pow(n/10%10,3) + pow(n%10,3) )             printf("%d\n",n);                   }        system("PAUSE");    return EXIT_SUCCESS;}
       
      
     

2-3 韩信点兵

#include 
     
      #include 
      
       int main(int argc, char *argv[]){    int n;    int a, b, c;    scanf("%d %d %d", &a, &b, &c);    for (n = 10; n<=100; n++)    {        if (n%3 == a && n%5 == b && n%7 == c)           {                printf("%d\n",n);                break;           }               }    if (n == 101) printf("No answer\n");        system("PAUSE");    return EXIT_SUCCESS;}
      
     

2-4 倒三角形

#include 
     
      #include 
      
       int main(int argc, char *argv[]){  int n;  int i, j, k;  scanf("%d", &n);  for( i = 1; i <= n; i++)      {          for ( k=1; k <= i-1; k++)              printf(" ");          for ( j=i; j <= 2*n-i; j++)          {               printf("#");          }                   printf("\n");      }  system("PAUSE");	  return 0;}
      
     

总结：printf((j<i || j > m-1-i) ? " " : "#"); 用这句来进行推断输出空格还是#，简单多了

2-6 调和级数

#include 
     
      #include 
      
       int main(int argc, char *argv[]){  int n;  double sum = 0;  scanf("%d",&n);    int i;  for(i=1; i<=n; i++)  {     sum += 1.0/i;       }    printf("%.3lf\n", sum);  system("PAUSE");	  return 0;}
      
     

2-7 近似计算

#include 
     
      #include 
      
       #include 
       
        int main(int argc, char *argv[]){  int i, j;  double sum = 0;  for( i = 1, j = 0; 1.0/i >= 1.0/pow(10,6); i += 2)  {     sum += (1.0/i)*pow(-1,j);          j++;  }    sum += (1.0/i)*pow(-1,j);  printf("%d\n", i);  printf("%.6lf\n", 4*sum);  system("PAUSE");	  return 0;}
       
      
     

总结： 能够用sign = -sign。来进行正负号的转换

2-8 子序列的和

#include 
     
      #include 
      
       #include 
       
        int main(int argc, char *argv[]){  int n, m;  scanf("%d %d", &n, &m);  double sum = 0.0;  int i;    for( i = n; i <= m; i++)  {     sum += 1.0/i/i;  }    printf("%.5lf\n", sum);  system("PAUSE");	  return 0;}
       
      
     

总结：仍然是怕溢出，1.0/i/i