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( )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!\007\n") ; 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=%.2lf\n", 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, "%d\n%d\n%.2lf\n", 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!\007\n") ; 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=%.2lf\n", 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, "%d\n%d\n%.2lf\n", 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!\007\n") ; 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=%.2lf\n", 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, "%d\n%d\n%.2lf\n", 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!\007\n") ; 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=%.2lf\n", 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, "%d\n%d\n%.2lf\n", 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!\007\n") ; 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=%.2lf\n", 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, "%d\n%d\n%.2lf\n", 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 第六套 =============================================================================== 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. 求这些数中的十位数位置上的数字是2、 4和8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 OUT6.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!\007\n") ; 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=%.2lf\n", 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("OUT6.DAT", "w") ; fprintf(fp, "%d\n%d\n%.2lf\n", 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 $OUT6.DAT 003 |160\|45\|5229.16 #E 第七套 =============================================================================== 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. 求这些数中的百位数位置上的数字是1、 5和7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 OUT7.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!\007\n") ; 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=%.2lf\n", 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("OUT7.DAT", "w") ; fprintf(fp, "%d\n%d\n%.2lf\n", 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 $OUT7.DAT 003 |160\|47\|5448.32 #E 第八套 =============================================================================== 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. 求这些数中的千位数位置上的数字大于个 位数位置上的数字的数的个数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OUT8.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!\007\n") ; 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=%.2lf\n", 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("OUT8.DAT", "w") ; fprintf(fp, "%d\n%d\n%.2lf\n", 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 $OUT8.DAT 003 |160\|81\|6617.44 #E 第九套 =============================================================================== 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. 求这些数中的百位数位置上的数字小于十 位数位置上的数字的数的个数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OUT9.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!\007\n") ; 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=%.2lf\n", 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("OUT9.DAT", "w") ; fprintf(fp, "%d\n%d\n%.2lf\n", 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 $OUT9.DAT 003 |160\|78\|5182.50 #E 第十套 =============================================================================== 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. 求这些数中的千位数位置上的数字与十位 数位置上的数字均为的奇数的个数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OUT10.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!\007\n") ; 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=%.2lf\n", 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("OUT10.DAT", "w") ; fprintf(fp, "%d\n%d\n%.2lf\n", 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 $OUT10.DAT 003 |160\|51\|5383.47 #E
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