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Not all corporations are traditional businesses that sell stock. ()
A. 所有的公司都不是出售股票的传统企业。
B. 并非所有的公司都是出售股票的传统企业。
C. 并非所有的公司都是非出售股票的传统企业。
参考答案
更多 “ Not all corporations are traditional businesses that sell stock. () A. 所有的公司都不是出售股票的传统企业。B. 并非所有的公司都是出售股票的传统企业。C. 并非所有的公司都是非出售股票的传统企业。 ” 相关考题
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已知在文件IN1.DAT中存有100个产品销售记录，每个产品销售记录由产品代码dm(字符型4位)、产品名称mc(字符型10位)、单价dj(整型)、数量s1(整型)、金额je(长整型)五部分组成。其中：金额＝单价×数量。函数ReadDat()的功能是读取这100个销售记录并存入结构数组.sell中。请编制函数SortDat()，其功能要求：按产品名称从小到大进行排列，若产品名称相同，则按金额从小到大进行排列，最终排列结果仍存入结构数组sell中，最后调用函数WriteDat()把结果输出到文件OUT1.DAT中。注意：部分源程序已给出。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。[试题程序]includeincludeincludeincludeincludedefine MAX 100typedef struct{ char dm[5]; /*产品代码* /char mc[11]; /*产品名称*/int dj; /*单价* /int si; /*数量*/long je; /*金额*/}PRO;PRO sell[-MAX];void ReadDat() ;void WriteDat() ;void SortDat(){}main(){memset(sell,0,sizeof(sell));ReadDat();SortDat();WriteDat();}void ReadDat(){FILE *fp;char str[80],ch[11];int i;fp=fopen("IN1.DAT","r");for (i=0;i＜100;i++){fgets(str,80,fp)memcpy(sell[i].dm,str,4)memcpy(sell[i].mc,str + 4,10);memcpy(ch,str + 14,4);ch[4]=0;sell[i].dj=atoi(ch)memcpy(ch,str +18,5);ch[5]=0;sell[i].s1=atoi(ch);sell[i].je=(long)sell[i].dj * sell[i].s1;}fclose(fp);}void WriteDat(){FILE * fp;int i;fp=fopen("OUT1.DAT","w");for(i=0;i100;i++)fprintf(fp,"%s%s%4d%5d%101d\n",sell[i].dm,sell[i].mc,sell[i].dj,sell[i].s1,sell[i].je)fclose(fp);}
考题
已知在文件IN9.DAT中存有100个产品销售记录，每个产品销售记录由产品代码dm(字符型4位)、产品名称mc(字符型10位)、单价dj(整型)、数量sl(整型)、金额je(长整型)五部分组成。其中，金额=单价X数量。函数ReadDat()的功能是读取这100个销售汜录并存入数组sell中。请编制函数SortDat()，其功能要求：按金额从小到大进行排列，若金额相同，则按产品代码从小到大进行排列，排列结果仍存入结构数组sell中，最后调用函数WriteDat()把结果输出到文件OUT9.DAT中。注意：部分源程序已给出。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。[试题程序]include＜stdio.h＞include＜memory.h＞include＜string.h＞include＜conio.h＞include＜stdlib.h＞define MAX 100typedef struct{char dm[5] /*产品代码*/char mc[11]； /*产品名称*/int dj； /*单价*/int sl； /*数量*/long je； /*金额*/}PRO；PR()sell[MAX];void ReadDat()；void WriteDat()；void SortDat(){}void main(){memset(sell,0,sizeof(sell))；ReadDat()；SortDat()；WriteDat()；}void ReadDat(){FILE*fp；char str[80],ch[ll]；int i；fp=fopen("IN9.DAT","r")；for(i=0；i＜100；i++){fgets(str,80,fp)；memcpy(sell[i].dm,str,4)；memcpy(sell[i].mc,str+4,10)；memcpy(ch.str+14,4)；ch[4]=0；sell[i].dj=atoi(ch)；memcpy(ch,str+18,5)；ch[5]=0；sell[i].sl=atoi(ch)；sell[i].je=(long)sell[i].dj*sell[i].sl；}fclosc(fp)；}void WriteDat(){FILE *fP；Int i；fp=fopen("OUT9.DAT","w")；for(i=0；i＜100；i++)fprintf(fp,"%s %s %4d %5d %101d＼n",sell[i].dm,sell[i].mc,sell[i].sl,sell[i].sl,sell[i].je)；fclose(fp)；}
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已知在文件IN. dat中存有100个产品销售记录，每个产品销售记录由产品代码code(字符型4位)、产品名称name(字符型10位)、单价uprice(整型)、数量amount(整型)、金额sum(长整型)5部分组成。其中：金额=单价×数量。函数Rdata()是读取这100个销售记录并存入结构数组sell中。请编写函数SortDat()，其功能要求：按金额从小到大进行排列，若金额相等，则按产品代码从小到大进行排列，最终排列结果仍存入结构数组sell中。最后调用函数Wdata()，把结果输出到OUT. dat文件中。注意：部分源程序已经给出。请勿改动主函数main()、读数据函数Rdata()和输出数据函数Wdata()的内容。include ＜stdio. h＞include ＜string. h＞include ＜conio. h＞include ＜stdlib. h＞define MAX 100typedef struct{ char code[5]; /* 产品代码 */char name[11]; /* 产品名称 */int uprice; /* 单价 */int amount; /* 数量 */long sum; /* 金额 */} PRO;PRO sell [MAX];void Rdata();void Wdata();void SortDat(){}void main (){ memset(sell, 0, sizeof(sell)Rdata();SortDat();Wdata();}void Rdata(){ FILE *fp;char str[80], ch[11];int i;fp = fopen("IN. dat", "r");for (i=0; i＜100; i++){ fgets(str, 80, fp);memcpy(sell[i].code, str, 4);memcpy(sell[i].name, str+4, 10);memcpy(ch, str+14, 4);ch[4] = 0;sell[i].uprice = atoi(ch);memcpy(ch, str+18, 5);ch[5] = 0;sell[i]. amount = atoi(ch);sell[i].sum = (long)sell[i]. uprice * sell[i]. amount;}fclose(fp);}void Wdata(){ FILE *fp;int i;fp = fopen("OUT. dat", "w");for (i=0; i＜100; i++){ printf("%s %s %4d %5d %5d\n", sell[i]. code, sell[i].name,sell[i].uprice, sell[i]. amount, sell[i]. sum);fprintf(fp, "%s %s %4d %5d %5d\n", sell[i]. code,sell[i]. name, sell[i]. uprice, sell[i]. amount, sell[i]. sum);}fclose(fp);}
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