C语言每日一练——第4题

一、题目要求

已知数据文件in.dat中有300个四位数,并调用readDat()函数把这些数存储数组a中,编写函数jsValue(),其功能是：求出所有这些四位数是素数的个数cnt,再把所有满足此条件的四位数依次存入数组b中,然后对数组b的按从小到大顺序排列,最后编写函数wirteDat()把数组b中的数输出到out.dat文件中。

二、in.dat文件内容

7704,9774,3213,5845,7703,2902,1925,2584,7429,6646
3793,8640,4364,3032,9190,2498,6394,3796,2434,6791
5552,9052,9159,8245,7965,1071,8157,6526,2164,6767
4853,1174,4564,5427,7279,9832,3640,7897,3211,4788
6556,4781,2168,6014,7795,6536,9461,4287,8029,8303
2609,8795,8386,3911,8130,7268,1906,7345,3678,5823
8074,6963,7874,6066,4509,8976,1740,2026,7047,4730
9538,2193,8473,6729,4207,5095,4623,4933,6226,8098
1277,5480,1906,4109,9047,1986,5882,8444,1029,6515
2612,7297,6934,8495,7731,1625,1305,7953,7385,3620
7206,4549,5736,3131,6875,6917,6087,4640,5314,8646
6077,5677,7345,4143,1513,8101,2038,9879,3946,6643
7763,6108,4788,9055,8186,5416,5699,6733,5416,2320
2430,4772,9962,9676,1319,7767,7369,1411,3663,2876
2245,4810,5400,6045,1504,4178,7264,2763,7050,1490
8790,4014,9280,7105,7113,3600,8221,8203,6127,7313
1850,5983,9689,4102,1036,6375,5064,7188,9887,9800
6871,4856,2432,9562,3426,9013,5104,4724,4785,9953
5939,6668,8700,4730,9737,4055,6544,9763,1034,2855
8078,5111,4817,4593,9876,1189,2507,1197,1782,4347
7961,8612,9699,6108,4221,2489,8895,3672,8226,8724
6302,7624,7506,8568,4013,2785,3528,6712,1155,3150
1388,5803,6328,6841,5825,7802,2022,6290,1775,3652
9699,6573,8082,9016,4369,4002,3205,2688,8414,7572
2830,1231,3888,5137,5555,8976,2543,6542,8986,5921
5346,1368,2566,3304,4089,3880,1244,6650,6523,6123
1243,7488,5262,9992,3115,9270,6989,1595,1283,2645
4075,9471,7983,9093,7642,8847,5686,4168,8550,8941
5188,2290,3219,2853,6978,9053,5843,7153,6382,6289
3022,5591,9981,7296,2421,7328,9575,7586,1190,8504

三、程序代码

#include <stdio.h>  
int a[300], b[300], cnt=0;  

void readDat()
{
  FILE *fp;  
  int i;  
  fp = fopen("in.dat","r");  
  for(i = 0; i < 300; i++)
  {
      fscanf(fp, "%d,", &a[i]);
  }
  fclose(fp);  
}

void writeDat()
{
  FILE *fp;  
  int i;   
  fp = fopen("out.dat","w");  
  fprintf(fp, "%d\n", cnt);  
  for(i = 0; i < cnt; i++)
  { 
      fprintf(fp, "%d\n", b[i]);
  }
  fclose(fp);  
}



 
int isP(int m)  
{
  int i;  
  for(i = 2; i < m; i++)  
  if(m % i == 0)
    return 0;  
  return 1;  
}
void jsValue()
{
    int i,j;
    int temp; 
    for(i=0;i<300;i++)
    {
        if(isP(a[i]))
        {
            b[cnt++]=a[i];
        }
    }
    
    //方法一：冒泡排序 
    for(i=0;i<cnt;i++)
    {
        for(j=0;j<cnt-i-1;j++)
        {    
            if(b[j]>b[j+1])
            {
                temp=b[j];
                b[j]=b[j+1];
                b[j+1]=temp;
            }
        }
    }
    //方法二：
    /*for(i = 0; i < cnt; i++)   
    {
        for(j = i + 1; j < cnt; j++)   
        {
            if(b[i] > b[j])   
            {
                temp = b[j];  
                b[j] = b[i];  
                b[i] = temp;  
            }
        }
    }*/
}
void main()
{
  int i;  
  readDat();  
  jsValue();  
  writeDat();  
  printf("cnt=%d\n", cnt);  
  for(i = 0; i < cnt; i++)
  {
      printf("b[%d]=%d\n", i, b[i]);
  }
}

四、程序运行结果