C语言每日一练——第2题

一、题目要求

已知数据文件in.dat中存有300个四位数,并调用读函数readDat()把这些数存入数组a中,请编制一函数jsValue(),其功能是:求出所有这些四位数是素数的个数cnt,再求出所有满足此条件的四位数平均值pjz1,以及所有不满足此条件的四位数平均数pjz2。最后main()函数调用写函数把结果cnt,pjz1,pjz2,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
View Code

三、程序代码

#include <stdio.h>  
int a[300], cnt=0;  
double pjz1=0.0, pjz2=0.0;  

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

//把运行结果输出到out.dat文件中
void writeDat()  
{
  FILE *fp;   
  fp = fopen("out.dat","w");  
  fprintf(fp, "%d\n%7.2lf\n%7.2lf\n", cnt, pjz1, pjz2);  
  fclose(fp);  
}

//判断是否是素数
int isP(int m)  
{
  int i;
  for(i = 2; i < m; i++)
  {
      if(m % i == 0)
      {
          return 0;
      }
  }
  return 1;
}
void jsValue() 
{
    int i=0,flag;
    int j=0,k=0;
    int sList[300];        //存放素数的数组
    int nsList[300];    //存放非素数的数组
    for(i=0;i<300;i++)
    {
        flag=isP(a[i]);
        if(flag)
        {
            cnt++;
            sList[j++]=a[i];
        }
        else
        {
            nsList[k++]=a[i];
        }
    }
    for(i=0;i<j;i++)
    {
        pjz1+=sList[i];

    }
    for(i=0;i<k;i++)
    {
        pjz2+=nsList[i];
    }
    pjz1/=j;
    pjz2/=k;

}
main()
{
  readDat();  
  jsValue();  
  writeDat();  
  printf("cnt=%d\n满足条件的平均值pzj1=%7.2lf\n不满足条件的平均值pzj2=%7.2lf\n", cnt,pjz1,pjz2);  
}
View Code

四、程序运行结果

 

posted @ 2018-09-04 23:01  深巷老猫  阅读(208)  评论(0编辑  收藏  举报