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2016年計(jì)算機(jī)四級(jí)上機(jī)編程試題及答案
第一套
===============================================================================
試題說(shuō)明 :
===============================================================================
已知在文件IN.DAT中存有若干個(gè)(個(gè)數(shù)<200)四位數(shù)字的正整數(shù), 函數(shù)ReadDat( )編制函數(shù)CalValue( ), 其功能要求: 1. 求出這文件中共有多少個(gè)正整數(shù)totNum; 2.求出這些數(shù)中的各位數(shù)字之和是奇數(shù)的數(shù)的個(gè)數(shù)totCnt, 以及滿足此條件的這些數(shù)的算術(shù)平均值totPjz, 最后調(diào)用函數(shù)WriteDat()把所求的結(jié)果輸出到文件OUT1.DAT中。
注意: 部分源程序存放在PROG1.C中。
請(qǐng)勿改動(dòng)主函數(shù)main( )、讀數(shù)據(jù)函數(shù)ReadDat()和輸出數(shù)據(jù)
函數(shù)WriteDat()的內(nèi)容。
===============================================================================
程序 :
===============================================================================
#include
#include
#define MAXNUM 200
int xx[MAXNUM] ;
int totNum = 0 ; /* 文件IN.DAT中共有多少個(gè)正整數(shù) */
int totCnt = 0 ; /* 符合條件的正整數(shù)的個(gè)數(shù) */
double totPjz = 0.0 ; /* 平均值 */
int ReadDat(void) ;
void WriteDat(void) ;
void CalValue(void)
{
}
void main()
{
clrscr() ;
if(ReadDat()) {
printf("數(shù)據(jù)文件IN.DAT不能打開(kāi)!\007\n") ;
return ;
}
CalValue() ;
printf("文件IN.DAT中共有正整數(shù)=%d個(gè)\n", totNum) ;
printf("符合條件的正整數(shù)的個(gè)數(shù)=%d個(gè)\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) ;
}
===============================================================================
所需數(shù)據(jù) :
===============================================================================
@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
===============================================================================
試題說(shuō)明 :
===============================================================================
已知在文件IN.DAT中存有若干個(gè)(個(gè)數(shù)<200)四位數(shù)字的正整數(shù), 函數(shù)ReadDat( )是讀取這若干個(gè)正整數(shù)并存入數(shù)組xx中。請(qǐng)編制函數(shù)CalValue( ), 其功能要求: 1. 求出這文件中共有多少個(gè)正整數(shù)totNum; 2.求出這些數(shù)中的各位數(shù)字之和是偶數(shù)的數(shù)的個(gè)數(shù)totCnt, 以及滿足此條件的這些數(shù)的算術(shù)平均值totPjz, 最后調(diào)用函數(shù)WriteDat()把所求的結(jié)果輸出到文件OUT2.DAT中。
注意: 部分源程序存放在PROG1.C中。
請(qǐng)勿改動(dòng)主函數(shù)main( )、讀數(shù)據(jù)函數(shù)ReadDat()和輸出數(shù)據(jù)
函數(shù)WriteDat()的內(nèi)容。
===============================================================================
程序 :
===============================================================================
#include
#include
#define MAXNUM 200
int xx[MAXNUM] ;
int totNum = 0 ; /* 文件IN.DAT中共有多少個(gè)正整數(shù) */
int totCnt = 0 ; /* 符合條件的正整數(shù)的個(gè)數(shù) */
double totPjz = 0.0 ; /* 平均值 */
int ReadDat(void) ;
void WriteDat(void) ;
void CalValue(void)
{
}
void main()
{
clrscr() ;
if(ReadDat()) {
printf("數(shù)據(jù)文件IN.DAT不能打開(kāi)!\007\n") ;
return ;
}
CalValue() ;
printf("文件IN.DAT中共有正整數(shù)=%d個(gè)\n", totNum) ;
printf("符合條件的正整數(shù)的個(gè)數(shù)=%d個(gè)\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) ;
}
===============================================================================
所需數(shù)據(jù) :
===============================================================================
@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
===============================================================================
試題說(shuō)明 :
===============================================================================
已知在文件IN.DAT中存有若干個(gè)(個(gè)數(shù)<200)四位數(shù)字的正整數(shù), 函數(shù)ReadDat( )是讀取這若干個(gè)正整數(shù)并存入數(shù)組xx中。請(qǐng)編制函數(shù)CalValue( ), 其功能要求: 1. 求出這文件中共有多少個(gè)正整數(shù)totNum; 2. 求這些數(shù)右移1位后, 產(chǎn)生的新數(shù)是奇數(shù)的數(shù)的個(gè)數(shù)totCnt, 以及滿足此條件的這些數(shù)(右移前的值)的算術(shù)平均值totPjz, 最后調(diào)用函數(shù)WriteDat()把所求的結(jié)果輸出到文件OUT3.DAT中。
注意: 部分源程序存放在PROG1.C中。
請(qǐng)勿改動(dòng)主函數(shù)main( )、讀數(shù)據(jù)函數(shù)ReadDat()和輸出數(shù)據(jù)
函數(shù)WriteDat()的內(nèi)容。
===============================================================================
程序 :
===============================================================================
#include
#include
#define MAXNUM 200
int xx[MAXNUM] ;
int totNum = 0 ; /* 文件IN.DAT中共有多少個(gè)正整數(shù) */
int totCnt = 0 ; /* 符合條件的正整數(shù)的個(gè)數(shù) */
double totPjz = 0.0 ; /* 平均值 */
int ReadDat(void) ;
void WriteDat(void) ;
void CalValue(void)
{
}
void main()
{
clrscr() ;
if(ReadDat()) {
printf("數(shù)據(jù)文件IN.DAT不能打開(kāi)!\007\n") ;
return ;
}
CalValue() ;
printf("文件IN.DAT中共有正整數(shù)=%d個(gè)\n", totNum) ;
printf("符合條件的正整數(shù)的個(gè)數(shù)=%d個(gè)\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) ;
}
===============================================================================
所需數(shù)據(jù) :
===============================================================================
@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
===============================================================================
試題說(shuō)明 :
===============================================================================
已知在文件IN.DAT中存有若干個(gè)(個(gè)數(shù)<200)四位數(shù)字的正整數(shù), 函數(shù)ReadDat( )是讀取這若干個(gè)正整數(shù)并存入數(shù)組xx中。請(qǐng)編制函數(shù)CalValue( ), 其功能要求: 1. 求出這文件中共有多少個(gè)正整數(shù)totNum; 2. 求這些數(shù)右移1位后, 產(chǎn)生的新數(shù)是偶數(shù)的數(shù)的個(gè)數(shù)totCnt, 以及滿足此條件的這些數(shù)(右移前的值)的算術(shù)平均值totPjz, 最后調(diào)用函數(shù)WriteDat()把所求的結(jié)果輸出到文件OUT4.DAT中。
注意: 部分源程序存放在PROG1.C中。
請(qǐng)勿改動(dòng)主函數(shù)main( )、讀數(shù)據(jù)函數(shù)ReadDat()和輸出數(shù)據(jù)
函數(shù)WriteDat()的內(nèi)容。
===============================================================================
程序 :
===============================================================================
#include
#include
#define MAXNUM 200
int xx[MAXNUM] ;
int totNum = 0 ; /* 文件IN.DAT中共有多少個(gè)正整數(shù) */
int totCnt = 0 ; /* 符合條件的正整數(shù)的個(gè)數(shù) */
double totPjz = 0.0 ; /* 平均值 */
int ReadDat(void) ;
void WriteDat(void) ;
void CalValue(void)
{
}
void main()
{
clrscr() ;
if(ReadDat()) {
printf("數(shù)據(jù)文件IN.DAT不能打開(kāi)!\007\n") ;
return ;
}
CalValue() ;
printf("文件IN.DAT中共有正整數(shù)=%d個(gè)\n", totNum) ;
printf("符合條件的正整數(shù)的個(gè)數(shù)=%d個(gè)\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) ;
}
===============================================================================
所需數(shù)據(jù) :
===============================================================================
@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
===============================================================================
試題說(shuō)明 :
===============================================================================
已知在文件IN.DAT中存有若干個(gè)(個(gè)數(shù)<200)四位數(shù)字的正整數(shù), 函數(shù)ReadDat( )是讀取這若干個(gè)正整數(shù)并存入數(shù)組xx中。請(qǐng)編制函數(shù)CalValue( ), 其功能要求: 1. 求出這文件中共有多少個(gè)正整數(shù)totNum; 2. 求這些數(shù)中的個(gè)位數(shù)位置上的數(shù)字是3、6和9的數(shù)的個(gè)數(shù)totCnt, 以及滿足此條件的這些數(shù)的算術(shù)平均值totPjz, 最后調(diào)用函數(shù)WriteDat( )把所求的結(jié)果輸出到文件OUT5.DAT中。
注意: 部分源程序存放在PROG1.C中。
請(qǐng)勿改動(dòng)主函數(shù)main( )、讀數(shù)據(jù)函數(shù)ReadDat()和輸出數(shù)據(jù)
函數(shù)WriteDat()的內(nèi)容。
===============================================================================
程序 :
===============================================================================
#include
#include
#define MAXNUM 200
int xx[MAXNUM] ;
int totNum = 0 ; /* 文件IN.DAT中共有多少個(gè)正整數(shù) */
int totCnt = 0 ; /* 符合條件的正整數(shù)的個(gè)數(shù) */
double totPjz = 0.0 ; /* 平均值 */
int ReadDat(void) ;
void WriteDat(void) ;
void CalValue(void)
{
}
void main()
{
clrscr() ;
if(ReadDat()) {
printf("數(shù)據(jù)文件IN.DAT不能打開(kāi)!\007\n") ;
return ;
}
CalValue() ;
printf("文件IN.DAT中共有正整數(shù)=%d個(gè)\n", totNum) ;
printf("符合條件的正整數(shù)的個(gè)數(shù)=%d個(gè)\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) ;
}
===============================================================================
所需數(shù)據(jù) :
===============================================================================
@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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