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全国计算机等级考试四级上机编程题型1

13-05-21

  第一套    ===============================================================================  试题说明 :  ==========================================================================...

  第一套
  
  ===============================================================================
  试题说明 :
  ===============================================================================
  已知在文件IN.DAT中存有若干个(个数<200)四位数字的正整数, 函数ReadDat( )编制函数CalValue( ), 其功能要求: 1. 求出这文件中共有多少个正整数totNum; 2.求出这些数中的各位数字之和是奇数的数的个数totCnt, 以及满足此条件的这些数的算术平均值totPjz, 最后调用函数WriteDat()把所求的结果输出到文件OUT1.DAT中。
  注意: 部分源程序存放在PROG1.C中。
  请勿改动主函数main( )、读数据函数ReadDat()和输出数据
  函数WriteDat()的内容。
  ===============================================================================
  程序 :
  ===============================================================================
  #include
  #include
  #define MAXNUM 200
  
  int xx[MAXNUM] ;
  int totNum = 0 ; /* 文件IN.DAT中共有多少个正整数 */
  int totCnt = 0 ; /* 符合条件的正整数的个数 */
  double totPjz = 0.0 ; /* 平均值 */
  int ReadDat(void) ;
  void WriteDat(void) ;
  void CalValue(void)
  {
  
  
  
  }
  
  void main()
  {
   clrscr() ;
   if(ReadDat()) {
   printf("数据文件IN.DAT不能打开!07n") ;
   return ;
   }
   CalValue() ;
   printf("文件IN.DAT中共有正整数=%d个n", totNum) ;
   printf("符合条件的正整数的个数=%d个n", totCnt) ;
   printf("平均值=%.2lfn", totPjz) ;
   WriteDat() ;
  }
  
  int ReadDat(void)
  {
   FILE *fp ;
   int i = 0 ;
  
   if((fp = fopen("in.dat", "r")) == NULL) return 1 ;
   while(!feof(fp)) {
   fscanf(fp, "%d,", &xx[i++]) ;
   }
   fclose(fp) ;
   return 0 ;
  }
  
  void WriteDat(void)
  {
   FILE *fp ;
  
   fp = fopen("OUT1.DAT", "w") ;
   fprintf(fp, "%dn%dn%.2lfn", totNum, totCnt, totPjz) ;
   fclose(fp) ;
  }
  ===============================================================================
  所需数据 :
  ===============================================================================
  @2 IN.DAT 016
  6045,6192,1885,3580,8544,6826,5493,8415,3132,5841,
  6561,3173,9157,2895,2851,6082,5510,9610,5398,5273,
  3438,1800,6364,6892,9591,3120,8813,2106,5505,1085,
  5835,7295,6131,9405,6756,2413,6274,9262,5728,2650,
  6266,5285,7703,1353,1510,2350,4325,4392,7573,8204,
  7358,6365,3135,9903,3055,3219,3955,7313,6206,1631,
  5869,5893,4569,1251,2542,5740,2073,9805,1189,7550,
  4362,6214,5680,8753,8443,3636,4495,9643,3782,5556,
  1018,9729,8588,2797,4321,4714,9658,8997,2080,5912,
  9968,5558,9311,7047,6138,7618,5448,1466,7075,2166,
  4025,3572,9605,1291,6027,2358,1911,2747,7068,1716,
  9661,5849,3210,2554,8604,8010,7947,3685,2945,4224,
  7014,9058,6259,9503,1615,1060,7787,8983,3822,2471,
  5146,7066,1029,1777,7788,2941,3538,2912,3096,7421,
  9175,6099,2930,4685,8465,8633,2628,7155,4307,9535,
  4274,2857,6829,6226,8268,9377,9415,9059,4872,6072,
  #E
  @3 $OUT1.DAT 003
  |160|69|5460.51
  #E
  第二套
  
  ===============================================================================
  试题说明 :
  ===============================================================================
   已知在文件IN.DAT中存有若干个(个数<200)四位数字的正整数, 函数ReadDat( )是读取这若干个正整数并存入数组xx中。请编制函数CalValue( ), 其功能要求: 1. 求出这文件中共有多少个正整数totNum; 2.求出这些数中的各位数字之和是偶数的数的个数totCnt, 以及满足此条件的这些数的算术平均值totPjz, 最后调用函数WriteDat()把所求的结果输出到文件OUT2.DAT中。
   注意: 部分源程序存放在PROG1.C中。
   请勿改动主函数main( )、读数据函数ReadDat()和输出数据
  函数WriteDat()的内容。
  ===============================================================================
  程序 :
  ===============================================================================
  #include
  #include
  #define MAXNUM 200
  
  int xx[MAXNUM] ;
  int totNum = 0 ; /* 文件IN.DAT中共有多少个正整数 */
  int totCnt = 0 ; /* 符合条件的正整数的个数 */
  double totPjz = 0.0 ; /* 平均值 */
  int ReadDat(void) ;
  void WriteDat(void) ;
  void CalValue(void)
  {
  
  
  
  }
  
  void main()
  {
   clrscr() ;
   if(ReadDat()) {
   printf("数据文件IN.DAT不能打开!07n") ;
   return ;
   }
   CalValue() ;
   printf("文件IN.DAT中共有正整数=%d个n", totNum) ;
   printf("符合条件的正整数的个数=%d个n", totCnt) ;
   printf("平均值=%.2lfn", totPjz) ;
   WriteDat() ;
  }
  
  int ReadDat(void)
  {
   FILE *fp ;
   int i = 0 ;
  
   if((fp = fopen("in.dat", "r")) == NULL) return 1 ;
   while(!feof(fp)) {
   fscanf(fp, "%d,", &xx[i++]) ;
   }
   fclose(fp) ;
   return 0 ;
  }
  
  void WriteDat(void)
  {
   FILE *fp ;
  
   fp = fopen("OUT2.DAT", "w") ;
   fprintf(fp, "%dn%dn%.2lfn", totNum, totCnt, totPjz) ;
   fclose(fp) ;
  }
  ===============================================================================
  所需数据 :
  ===============================================================================
  @2 IN.DAT 016
  6045,6192,1885,3580,8544,6826,5493,8415,3132,5841,
  6561,3173,9157,2895,2851,6082,5510,9610,5398,5273,
  3438,1800,6364,6892,9591,3120,8813,2106,5505,1085,
  5835,7295,6131,9405,6756,2413,6274,9262,5728,2650,
  6266,5285,7703,1353,1510,2350,4325,4392,7573,8204,
  7358,6365,3135,9903,3055,3219,3955,7313,6206,1631,
  5869,5893,4569,1251,2542,5740,2073,9805,1189,7550,
  4362,6214,5680,8753,8443,3636,4495,9643,3782,5556,
  1018,9729,8588,2797,4321,4714,9658,8997,2080,5912,
  9968,5558,9311,7047,6138,7618,5448,1466,7075,2166,
  4025,3572,9605,1291,6027,2358,1911,2747,7068,1716,
  9661,5849,3210,2554,8604,8010,7947,3685,2945,4224,
  7014,9058,6259,9503,1615,1060,7787,8983,3822,2471,
  5146,7066,1029,1777,7788,2941,3538,2912,3096,7421,
  9175,6099,2930,4685,8465,8633,2628,7155,4307,9535,
  4274,2857,6829,6226,8268,9377,9415,9059,4872,6072,
  #E
  @3 $OUT2.DAT 003
  |160|91|5517.16
  #E   
  第三套
  
  ===============================================================================
  试题说明 :
  ===============================================================================
   已知在文件IN.DAT中存有若干个(个数<200)四位数字的正整数, 函数ReadDat( )是读取这若干个正整数并存入数组xx中。请编制函数CalValue( ), 其功能要求: 1. 求出这文件中共有多少个正整数totNum; 2. 求这些数右移1位后, 产生的新数是奇数的数的个数totCnt, 以及满足此条件的这些数(右移前的值)的算术平均值totPjz, 最后调用函数WriteDat()把所求的结果输出到文件OUT3.DAT中。
   注意: 部分源程序存放在PROG1.C中。
   请勿改动主函数main( )、读数据函数ReadDat()和输出数据
  函数WriteDat()的内容。
  ===============================================================================
  程序 :
  ===============================================================================
  #include
  #include
  #define MAXNUM 200
  
  int xx[MAXNUM] ;
  int totNum = 0 ; /* 文件IN.DAT中共有多少个正整数 */
  int totCnt = 0 ; /* 符合条件的正整数的个数 */
  double totPjz = 0.0 ; /* 平均值 */
  int ReadDat(void) ;
  void WriteDat(void) ;
  void CalValue(void)
  {
  
  
  
  }
  
  void main()
  {
   clrscr() ;
   if(ReadDat()) {
   printf("数据文件IN.DAT不能打开!07n") ;
   return ;
   }
   CalValue() ;
   printf("文件IN.DAT中共有正整数=%d个n", totNum) ;
   printf("符合条件的正整数的个数=%d个n", totCnt) ;
   printf("平均值=%.2lfn", totPjz) ;
   WriteDat() ;
  }
  
  int ReadDat(void)
  {
   FILE *fp ;
   int i = 0 ;
  
   if((fp = fopen("in.dat", "r")) == NULL) return 1 ;
   while(!feof(fp)) {
   fscanf(fp, "%d,", &xx[i++]) ;
   }
   fclose(fp) ;
   return 0 ;
  }
  
  void WriteDat(void)
  {
   FILE *fp ;
  
   fp = fopen("OUT3.DAT", "w") ;
   fprintf(fp, "%dn%dn%.2lfn", totNum, totCnt, totPjz) ;
   fclose(fp) ;
  }
  ===============================================================================
  所需数据 :
  ===============================================================================
  @2 IN.DAT 016
  6045,6192,1885,3580,8544,6826,5493,8415,3132,5841,
  6561,3173,9157,2895,2851,6082,5510,9610,5398,5273,
  3438,1800,6364,6892,9591,3120,8813,2106,5505,1085,
  5835,7295,6131,9405,6756,2413,6274,9262,5728,2650,
  6266,5285,7703,1353,1510,2350,4325,4392,7573,8204,
  7358,6365,3135,9903,3055,3219,3955,7313,6206,1631,
  5869,5893,4569,1251,2542,5740,2073,9805,1189,7550,
  4362,6214,5680,8753,8443,3636,4495,9643,3782,5556,
  1018,9729,8588,2797,4321,4714,9658,8997,2080,5912,
  9968,5558,9311,7047,6138,7618,5448,1466,7075,2166,
  4025,3572,9605,1291,6027,2358,1911,2747,7068,1716,
  9661,5849,3210,2554,8604,8010,7947,3685,2945,4224,
  7014,9058,6259,9503,1615,1060,7787,8983,3822,2471,
  5146,7066,1029,1777,7788,2941,3538,2912,3096,7421,
  9175,6099,2930,4685,8465,8633,2628,7155,4307,9535,
  4274,2857,6829,6226,8268,9377,9415,9059,4872,6072,
  #E
  @3 $OUT3.DAT 003
  |160|80|5537.54
  #E  第四套
  
  ===============================================================================
  试题说明 :
  ===============================================================================
   已知在文件IN.DAT中存有若干个(个数<200)四位数字的正整数, 函数ReadDat( )是读取这若干个正整数并存入数组xx中。请编制函数CalValue( ), 其功能要求: 1. 求出这文件中共有多少个正整数totNum; 2. 求这些数右移1位后, 产生的新数是偶数的数的个数totCnt, 以及满足此条件的这些数(右移前的值)的算术平均值totPjz, 最后调用函数WriteDat()把所求的结果输出到文件OUT4.DAT中。
   注意: 部分源程序存放在PROG1.C中。
   请勿改动主函数main( )、读数据函数ReadDat()和输出数据
  函数WriteDat()的内容。
  ===============================================================================
  程序 :
  ===============================================================================
  #include
  #include
  #define MAXNUM 200
  
  int xx[MAXNUM] ;
  int totNum = 0 ; /* 文件IN.DAT中共有多少个正整数 */
  int totCnt = 0 ; /* 符合条件的正整数的个数 */
  double totPjz = 0.0 ; /* 平均值 */  
  int ReadDat(void) ;
  void WriteDat(void) ;  
  void CalValue(void)
  {
  
  
  
  }
  
  void main()
  {
   clrscr() ;
   if(ReadDat()) {
   printf("数据文件IN.DAT不能打开!07n") ;
   return ;
   }
   CalValue() ;
   printf("文件IN.DAT中共有正整数=%d个n", totNum) ;
   printf("符合条件的正整数的个数=%d个n", totCnt) ;
   printf("平均值=%.2lfn", totPjz) ;
   WriteDat() ;
  }
  
  int ReadDat(void)
  {
   FILE *fp ;
   int i = 0 ;
  
   if((fp = fopen("in.dat", "r")) == NULL) return 1 ;
   while(!feof(fp)) {
   fscanf(fp, "%d,", &xx[i++]) ;
   }
   fclose(fp) ;
   return 0 ;
  }
  
  void WriteDat(void)
  {
   FILE *fp ;
  
   fp = fopen("OUT4.DAT", "w") ;
   fprintf(fp, "%dn%dn%.2lfn", totNum, totCnt, totPjz) ;
   fclose(fp) ;
  }
  ===============================================================================
  所需数据 :
  ===============================================================================
  @2 IN.DAT 016
  6045,6192,1885,3580,8544,6826,5493,8415,3132,5841,
  6561,3173,9157,2895,2851,6082,5510,9610,5398,5273,
  3438,1800,6364,6892,9591,3120,8813,2106,5505,1085,
  5835,7295,6131,9405,6756,2413,6274,9262,5728,2650,
  6266,5285,7703,1353,1510,2350,4325,4392,7573,8204,
  7358,6365,3135,9903,3055,3219,3955,7313,6206,1631,
  5869,5893,4569,1251,2542,5740,2073,9805,1189,7550,
  4362,6214,5680,8753,8443,3636,4495,9643,3782,5556,
  1018,9729,8588,2797,4321,4714,9658,8997,2080,5912,
  9968,5558,9311,7047,6138,7618,5448,1466,7075,2166,
  4025,3572,9605,1291,6027,2358,1911,2747,7068,1716,
  9661,5849,3210,2554,8604,8010,7947,3685,2945,4224,
  7014,9058,6259,9503,1615,1060,7787,8983,3822,2471,
  5146,7066,1029,1777,7788,2941,3538,2912,3096,7421,
  9175,6099,2930,4685,8465,8633,2628,7155,4307,9535,
  4274,2857,6829,6226,8268,9377,9415,9059,4872,6072,
  #E
  @3 $OUT4.DAT 003
  |160|80|5447.93
  #E   第五套
  
  ===============================================================================
  试题说明 :
  ===============================================================================
   已知在文件IN.DAT中存有若干个(个数<200)四位数字的正整数, 函数ReadDat( )是读取这若干个正整数并存入数组xx中。请编制函数CalValue( ), 其功能要求: 1. 求出这文件中共有多少个正整数totNum; 2. 求这些数中的个位数位置上的数字是3、6和9的数的个数totCnt, 以及满足此条件的这些数的算术平均值totPjz, 最后调用函数WriteDat( )把所求的结果输出到文件OUT5.DAT中。
   注意: 部分源程序存放在PROG1.C中。
   请勿改动主函数main( )、读数据函数ReadDat()和输出数据
  函数WriteDat()的内容。
  ===============================================================================
  程序 :
  ===============================================================================
  #include
  #include
  #define MAXNUM 200
  
  int xx[MAXNUM] ;
  int totNum = 0 ; /* 文件IN.DAT中共有多少个正整数 */
  int totCnt = 0 ; /* 符合条件的正整数的个数 */
  double totPjz = 0.0 ; /* 平均值 */  
  int ReadDat(void) ;
  void WriteDat(void) ;  
  void CalValue(void)
  {
  
  
  
  }
  
  void main()
  {
   clrscr() ;
   if(ReadDat()) {
   printf("数据文件IN.DAT不能打开!07n") ;
   return ;
   }
   CalValue() ;
   printf("文件IN.DAT中共有正整数=%d个n", totNum) ;
   printf("符合条件的正整数的个数=%d个n", totCnt) ;
   printf("平均值=%.2lfn", totPjz) ;
   WriteDat() ;
  }
  
  int ReadDat(void)
  {
   FILE *fp ;
   int i = 0 ;
  
   if((fp = fopen("in.dat", "r")) == NULL) return 1 ;
   while(!feof(fp)) {
   fscanf(fp, "%d,", &xx[i++]) ;
   }
   fclose(fp) ;
   return 0 ;
  }
  
  void WriteDat(void)
  {
   FILE *fp ;
  
   fp = fopen("OUT5.DAT", "w") ;
   fprintf(fp, "%dn%dn%.2lfn", totNum, totCnt, totPjz) ;
   fclose(fp) ;
  }
  ===============================================================================
  所需数据 :
  ===============================================================================
  @2 IN.DAT 016
  6045,6192,1885,3580,8544,6826,5493,8415,3132,5841,
  6561,3173,9157,2895,2851,6082,5510,9610,5398,5273,
  3438,1800,6364,6892,9591,3120,8813,2106,5505,1085,
  5835,7295,6131,9405,6756,2413,6274,9262,5728,2650,
  6266,5285,7703,1353,1510,2350,4325,4392,7573,8204,
  7358,6365,3135,9903,3055,3219,3955,7313,6206,1631,
  5869,5893,4569,1251,2542,5740,2073,9805,1189,7550,
  4362,6214,5680,8753,8443,3636,4495,9643,3782,5556,
  1018,9729,8588,2797,4321,4714,9658,8997,2080,5912,
  9968,5558,9311,7047,6138,7618,5448,1466,7075,2166,
  4025,3572,9605,1291,6027,2358,1911,2747,7068,1716,
  9661,5849,3210,2554,8604,8010,7947,3685,2945,4224,
  7014,9058,6259,9503,1615,1060,7787,8983,3822,2471,
  5146,7066,1029,1777,7788,2941,3538,2912,3096,7421,
  9175,6099,2930,4685,8465,8633,2628,7155,4307,9535,
  4274,2857,6829,6226,8268,9377,9415,9059,4872,6072,
  #E
  @3 $OUT5.DAT 003
  |160|43|5694.58
  #E
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