习题参考答案_2011_03_OUC_13_C语言程序设计项目化教程_武桂力_2....doc

13项目9 文件让计算机处理数据流参 考 答 案项目1一、选择题1D2B3B4A5A6C7B8A9C10A11B12C13D14A15B16C二判断题1错2错3对4对5对6对7错8对9错10对11对12对13对14对15错16对17对18对项目2一、选择题1D2A3A4C5D6A7C8B9B10C11A12C13A14B15B二、判断题1对2错3错4错5错6错7错8对9错10错11对12错13错14错15错项目3一、选择题1D2A3C4D5D6B7B8C9C10B11B12C13A14D15B16B17B18C19B二、判断题1错2对3错4对5对6错7错8对9错项目4一、选择题1C2B3C4C5A6D7B8A9C10C11B12A13D14D15B16B17B二、判断题1对2对3对4对5错6错7对8错9错10错11对12错13错14对15错16对17对18对19错项目5一、选择题1C2D3D4A5C6C7C8B9B10D11D12B13C14A15C16D17D18B二、判断题1对2错3错4错5对6对7对8错三、填空题10；4；2&ai；i%4=0；Printf(n)；3i=j,aij4strcmp(str0,str1)0?str0:str1s项目6一、选择题1C2A3C4B5B6D7A8A9C10A11D12D13A二、判断题1对2对3对4对5对6对7错8对9错10对11错12对13对14错15错16错三、填空题1程序中的main()函数2函数说明、函数体3max is242*i+1 a(i)a(i)5age(n-1)+2；age(5)65 10 97i10arrayiaverage(score)8i=3j=i-1ai-1j-19123三、编程题1int isprime(int a)int i;for(i=2;isqrt(double)a);i+)if(a%i=0) return 0;return 1;2int sum(int n)int i,k=0;for(i=0;i=n;i+) k+=i;return k;3float p(int n,int x)float t,t1,t2;if(n=0) return 1;else if(n=1) return x;elset1=(2*n-1)*x*p(n-1),x);t2=(n-1)*p(n-2),x);t=(t1-t2)/n;return t;4int max_value(int arr4)int i,j,max;max=arr00;for(i=0;i2;i+)for(j=0;jmax) max=arrij;return max;5#define M 50void main()int aM,c5=0,i,n,x;n=0;printf(Enter 1 or 2 or 3 or 4,to end with-1n);scanf(%d,&x);while(x!=-1)if(x=1 & x=4) an=x; n+;scanf(%d,&x);f(a,c,n);printf(Output the result: n;for(i=1;i=4;i+) printf(%d： %dn,i,ci);printf(n);f(int a,int c,int n)int i;for(i=0;in;i+) cai+;项目7一、选择题1B2D3D4C5B6C7D8B9D10A11B二、填空题1num=*b; num=*c; 2、hELLO! 3、ef 44 12 58 10 6* (a+i)=* (a+j);* (a+j)=t7二维数组各元素的五种表示方法：1) aij2) (*(a+i)j3) *(ai+j)4) *(*(a+i)+j)5) *(&a00+4*i+j)三、程序设计1main() int x,y,z,t ;int *p1,*p2,*p3;printf(“Please input 3 numbers:”);scanf(“%d,%d,%d”,&x,&y,&z);p1=&x;p2=&y;p3=&z;printf(“old values are :n”);printf(“%d%d%dn”,x,y,z);t=*p3;*p3=*p2;*p2=*p1;*p1=t;printf(“new valies are:n”);printf(“%d%d%d n”,x,y,z);2main() int *p,a20,i; printf(“Please input 10 numbersn”); for(i=0;i10;i+) scanf(“%d”,&ai); printf(“The original array is:”); for(p=data;pdata+20;p+) if(p-&data0)%5=0) printf(“n”); printf(“%4d”,*p); sort(data,20); printf(“the present array is:”); for(p=data;pdata+20;p+) if(p-&data0%5= =0) printf(“n”); printf(“%4d”,*p); void sort(int array,int n) int *p1,*p2,t; for(p1=array; p1array+(n-1);p1+) for(p2=p1+1;p2*p2) t=*p1; *p1=*p2; *p2=t; 3void inv(int *x,int n) int *p,temp,*i,*j,m=(n-1)/2; i=x;j=x+n-1;p=x+m; for(;i=p;i+,j-)temp=*i;*i=*j;*j=temp; return;main()int i,a10=3,7,9,11,0,6,7,5,4,2; printf(The original array:n); for(i=0;i10;i+) printf(%d,ai); printf(n); inv(a,10); printf(The array has benn inverted:n); for(i=0;i10;i+) printf(%d,ai); printf(n);4指针法：main()int a5,i,*pa;pa=a;for(i=0;i5;i+)*pa=i;pa+;pa=a;for(i=0;i5;i+)printf(a%d=%dn,i,*pa);pa+;下标法：main()int a5,i;for(i=0;i5;i+)ai=i;printf(a%d=%dn,i,ai);printf(n);5main() void sort(char *a ,int n) static char *name=“Changhua”, “Liping”, “Chenmei”, “Gaofeng” ; int n=4,i; sort(name,n); for(i=0;i5;i+) printf(%sn”,namei); void sort(char *a,int n) char *temp; int i,j; for(i=0;in-1;i+) for(j=0;j0) temp=namej; namej=namej+1; namej+1=temp; 项目8一、选择题1A2D3A4B5C6C二、填空题110，x212，035，34pperson+3，old=page; qname,old；5while(p!=NULL) c=c+1; p=pnext;项目9选择题1A2A3C4D5C6C7B8D二、程序设计1#includemain()FILE *fp;char ch;if(fp=fopen(text1.c,rt)=NULL)printf(Cannot open file strike any key exit!);getch();exit(1);ch=fgetc(fp);while (ch!=EOF)putchar(ch);ch=fgetc(fp);fclose(fp);2/* 函数功能： 把srcName文件内容拷贝到dstName 函数入口参数： 文件路径 函数返回值： 非0值表示拷贝成功，否则出错*/int CopyFile(const char* srcName, const char* dstName)#define BUF_SIZE 1024char bufBUF_SIZE;int fhSrc = -1;int fhDst = -1; int rval=1;int rtn;/* 打开文件 */fhSrc = open(srcName, O_RDONLY | O_BINARY);if (fhSrc = -1)goto ERROR;fhDst = open(dstName, O_WRONLY | O_CREAT | O_TRUNC | O_BINARY);if (fhDst = -1)goto ERROR;/* 拷贝文件 */while(rtn = read(fhSrc, buf, BUF_SIZE) 0)if (write(fhDst, buf, rtn) = -1)goto ERROR;if (rtn = 0)goto EXIT;ERROR:rval=0;EXIT:if (fhSrc != -1)close(fhSrc);if (fhDst != -1)close(fhDst);#undef BUF_SIZEreturn rval;3#include #include #include #include #define DEAL50/*设计最大交易次数*/struct dealstruct date dt;/*每笔交易的日期*/struct time ti;/*每笔交易的时间*/double earning;/*每笔交易的收入额度*/double payout;/*每笔交易的支出额度*/;typedef struct deal FINANCE;FINANCE personDEAL=0;float SumOfEarning(FINANCE per,int year)int i;float sum=0.0;for(i=0;iDEAL;i+)if(per.dt.da_year = year)sum += per.earning;return sum;float SumOfPayout(FINANCE per,int year)int i;float sum=0.0;for(i=0;iDEAL;i+) if(per.dt.da_year = year)sum += per.payout;return sum;float Balance(FINANCE *per)int i;float sum1=0.0,sum2=0.0;for(i=0;iearning;sum2 += (per+i)-payout;return sum1+sum2;void OneYearBalance(FINANCE *per,int year)int i;float sum1=0.0,sum2=0.0;for(i=0;idt.da_year != year)continue;sum1 += (per+i)-earning;sum2 += (per+i)-payout;printf(param of %d : ,year);printf(Sum of earning Sum of payout Balancen);printf(%28.2f%21.2f%19.2fn,sum1,sum2,sum1+sum2);void PrintBalance(FINANCE *per)int i;printf( Date Time Earning Payout Balance n);for(i=0;idt.da_year!=0)printf(%d/%d/%d ,(per+i)-dt.da_year,(per+i)-dt.da_mon,(per+i)-dt.da_day);printf(%d:%d:%d :,(per+i)-ti.ti_hour,(per+i)-ti.ti_min,(per+i)-ti.ti_sec);if(per+i)-earning!=0)printf(%8.2fn,(per+i)-earning);if(per+i)-payout!=0)printf( %8.2fn,(per+i)-payout);printf( %8.2fn,Balance(per);void PrintOneYear(FINANCE *per)int i,year;printf(Please Input one year :);scanf(%d,&year);printf( Date Time Earning Payout Balance n);for(i=0;idt.da_year != year) continue;if(per+i)-dt.da_year!=0)printf(%d/%d/%d ,(per+i)-dt.da_year,(per+i)-dt.da_mon,(per+i)-dt.da_day);printf(%d:%d:%d :,(per+i)-ti.ti_hour,(per+i)-ti.ti_min,(per+i)-ti.ti_sec);if(per+i)-earning!=0)printf(%8.2fn,(per+i)-earning);if(per+i)-payout!=0)printf( %8.2fn,(per+i)-payout);OneYearBalance(per,year);void GetDateTime(FINANCE *per)printf(Please input one deal:n);getdate(&per-dt);printf(%d/%d/%d :,per-dt.da_year,per-dt.da_mon,per-dt.da_day);gettime(&per-ti);printf(%d:%d:%d :,per-ti.ti_hour,per-ti.ti_min,per-ti.ti_sec);char InputOneDeal(FINANCE *per)char string10;printf(Please input deal (+/-)n);GetDateTime(per);scanf(%s,&string);if(string0=-)per-payout = atof(string);elseper-earning = atof(string);void Menu()printf(1.Input One dealn);printf(2.Print All the Balancen);printf(3.Print One year Balancen);printf(4.Exi