一、題目要求

已知數據文件in.dat中有300個四位數,並調用readDat()函數把這些數存儲數組a中,編寫函數jsValue(),其功能是：求出所有這些四位數是素數的個數cnt,再把所有滿足此條件的四位數依次存入數組b中,然後對數組b的按從小到大順序排列,最後編寫函數wirteDat()把數組b中的數輸出到out.dat文件中。

二、in.dat文件內容

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、程序代碼

#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、程序運行結果

C語言每日一練——第四題