2013省考上机模拟两个叠加-赵功建.doc

模拟01#include void main() int i,j,flag,n; n=0; i=7; while(n=10) /*$ERROR1$*/ flag=0; for(j=2;j=0) /*$ERROR3$*/ n=n+1; printf(%dn,i); i=i+10; $ERROR1$：while(n=10)$ERROR2$：if(i%j=0)$ERROR3$：if(flag=0)#include void main() int score; printf(Enter your score:); scanf(%d,_); /*$BLANK1$*/ printf(grade is:); switch(_) /*$BLANK2$*/ case 10: case 9:printf(An);break; case 8:printf(Bn);break; case 7:printf(Cn);break; case 6:printf(Dn);break; _:printf(En); /*$BLANK3$*/ $BLANK1$：&score$BLANK2$：score/10$BLANK3$：default#include #include void PRINT(double s) FILE *out; printf(s=%.2fn,s); if(out=fopen(C:24000101RESULT.DAT,w+)!=NULL) fprintf(out,s=%.2f,s); fclose(out);void main() double s=0; int i; for(i=300;i=2013;i+)if(i%9=0|i%13=0) s=s+sqrt(i); PRINT(s);模拟02#include void main() int a; scanf(%d,a); /*$ERROR1$*/ if(a100) printf(Data Error!n); else else if(a=90) printf(An); /*$ERROR2$*/ else if(a=80) printf(Bn); else if(a=70) printf(Cn); else if(a=60) printf(Dn); printf(En); /*$ERROR3$*/ $ERROR1$：scanf(%d,&a);$ERROR2$：if(a=90) printf(An);(去掉开始的else)$ERROR3$：else printf(En);(行首添加else)#include void main()long s, t;printf(nPlease enter s:); scanf(%ld,&s);t = s%10; s/= 10;while(_) /*$BLANK1$*/ t =_+ s%10;/*$BLANK2$*/s/= 10;printf(The result is: _n, t); /*$BLANK3$*/$BLANK1$：s0$BLANK2$：t*10$BLANK3$：%ld#include void PRINT(int a,int b,int c) FILE *out; printf(big=%d,mid=%d,small=%dn,a,b,c); if(out=fopen(C:24000102RESULT.DAT,w+)!=NULL) fprintf(out,big=%dp,mid=%dp,small=%dp,a,b,c); fclose(out);void main() int a,b,c; for(a=1;a25;a+)for(b=1;b33;b+) for(c=5;c100;c=c+5) if(a+b+c=100&4*a+3*b+c/5=100)PRINT(a,b,c);模拟03#include void main()int i,j;int n;scanf(%d,n);/*$ERROR1$*/for (i=1;i=n;i+)for (j=1;j=n-i;j+)printf( );j=0;/*$ERROR2$*/while (j=i)printf(%-2d,i);j+;print(n)/*$ERROR3$*/$ERROR1$：scanf(%d,&n);$ERROR2$：j=1;$ERROR3$：printf(n);#include void main() int i,a,b,c,n; n=_; /*$BLANK1$*/ for(i=101;i=999;i+) a=i%10; b=i/10%10; c=_; /*$BLANK2$*/ if(a*a*a+b*b*b+c*c*c=i) printf(%6d,i); _; /*$BLANK3$*/ printf(nn=%dn,n);$BLANK1$：0$BLANK2$：i/100$BLANK3$：n+#include void PRINT(long s) FILE *out; printf(s=%ldn,s); if(out=fopen(C:24000103RESULT.DAT,w+)!=NULL) fprintf(out,s=%ld,s); fclose(out);void main() long s=0;int i,j,k,t;for(i=0;i=3;i+) for(j=0;j=3;j+) for(k=0;k=3;k+) t=i*100+j*10+k; if(t%2=1) s=s+t; PRINT(s);模拟04输出一维数组中的最大元素及其下标值。#include void main() int a10=-3,1,-5,4,9,0,-8,7,-6,2; int i,addr; addr=1; /*$ERROR1$*/ i=1; while(i=10) /*$ERROR2$*/ if(aaddrai) i=addr; /*$ERROR3$*/ i+; printf(max=%d,address=%dn,aaddr,addr);$ERROR1$：addr=0;$ERROR2$：while(i10)$ERROR3$：addr=i;#include #include void main()char s100;int left,right,middle,lenth;printf(Please input a string:);gets(s);lenth = strlen(s);left = _; /*$BLANK1$*/middle = lenth/2;while(leftmiddle)right = lenth-1-left;if(sleft _ sright) /*$BLANK2$*/ left+; elsebreak;if(left _ middle) /*$BLANK3$*/printf(Non);elseprintf(Yesn); $BLANK1$：left=0;$BLANK2$：if(sleft = sright)$BLANK3$：if(leftmiddle)#include void PRINT(double s) FILE *out; printf(s=%.2fn,s); if(out=fopen(C:24000104RESULT.DAT,w+)!=NULL) fprintf(out,s=%.2f,s); fclose(out);void main() double s=100,h=100; int i; for(i=1;i8;i+) h=h/2; s=s+2*h; PRINT(s);模拟05#include void main()char s81;int num5,i,k,j;printf(Please enter a string:);gets(s);i=5;for (k=0;k=0) numi+;j-;/*$ERROR3$*/for (i=0;i5;i+)printf(%4d,numi);printf(n);$ERROR1$：numk=0;$ERROR2$：switch(sj)$ERROR3$：j+;#include void main() int i,j,row,col; int a34=3,5,1,8,6,4,11,7,9,3,10,2; row=col=0; for(i=0;i_;i+) /*$BLANK1$*/ for(j=0;j4;j+) if(arowcol _ ) /*$BLANK2$*/ row=i; col=j; printf(max=%d,row=%d,col=%dn,_,row,col); /*$BLANK3$*/$BLANK1$：3$BLANK2$：aij$BLANK3$：arowcol#include void PRINT(int n) FILE *out; printf(n=%dn,n); if(out=fopen(C:24000105RESULT.DAT,w+)!=NULL) fprintf(out,n=%d,n); fclose(out);void main()int n=0,s=10000; while(s=1000) n+; s=s-n; PRINT(n);模拟06#include struct student int num; char name12; int score;void main() student stud6= 1001,Pan Dong,48, /*$ERROR1$*/ 1002,Zhao Hua,62, 1003,Hu Litai,93, 1004,Zhang Li,85, 1005,Liu Ming,58, 1006,Xin Peng,37; int i,n=0; printf(numtnamettscoren); i=1;/*$ERROR2$*/ while(i6) if(studi.score60) printf(%dt%st%dn,studi.num,,studi.score); n-; /*$ERROR3$*/ i+; printf(n=%dn,n);$ERROR1$：struct student stud6= 1001,Pan Dong,48,$ERROR2$：i=0;$ERROR3$：n+;#include void main() int i; float a9,max,min,ave; printf(Input array:n); for(i=0;i9;i+) scanf(%f,_); /*$BLANK1$*/ max=min=ave=_; /*$BLANK2$*/ for(i=1;i9;i+) ave+=ai; if(maxai) _; /*$BLANK3$*/ ave=(ave-max-min)/7; printf(Mark=%.3fn,ave);$BLANK1$：&ai$BLANK2$：a0$BLANK3$：min=ai#include void PRINT(int count) FILE *out; printf(count=%dn,count); if(out=fopen(C:24000106RESULT.DAT,w+)!=NULL) fprintf(out,count=%d,count); fclose(out);void main()int count=0; int a,b,c; for(a=1;a100;a+) for(b=1;b20;b+) for(c=1;c10;c+) if(a+5*b+10*c=100) count+; PRINT(count);模拟07#include void main() char s80; int i; getc(s); /*$ERROR1$*/ for(i=0; si!=0; i+) if(si=z) /*$ERROR2$*/ si=a; else if(si=a&si=y) si=si+1; printf(%cn,s); /*$ERROR3$*/$ERROR1$：gets(s);$ERROR2$：if(si=z)$ERROR3$：printf(%sn,s);#include void main()double x,y;printf(Please input x:);scanf(%lf,_); /*$BLANK1$*/if(x_3) /*$BLANK2$*/y=8;else if(x_50) /*$BLANK3$*/ y=2*(8+(x-3)*1.5);else y=8+(x-3)*1.5; printf(y=%.1fn,y);$BLANK1$：&x$BLANK2$：=#include void PRINT(double s) FILE *out; printf(s=%.5fn,s); if(out=fopen(C:24000107RESULT.DAT,w+)!=NULL) fprintf(out,s=%.5f,s); fclose(out);void main()double s=0,f; int x; for(x=-10;x=10;x+) if(x=0|x=2) f=0;else if(x0) f=1.0*(x-1)/(x-2); else f=1.0*(x+1)/(x-2);s=s+f; PRINT(s);模拟08#include double fun(int n) int i; double s; s=1.0; for(i=2;i=n;i+) s+=1.0/i; return ; /*$ERROR1$*/void main() int n; double result; printf(nplease enter 1 integer numbers:n); scanf(%d,n); /*$ERROR2$*/ result=fun(&n); /*$ERROR3$*/ printf(nthe result is %lfn,result); $ERROR1$：return s;$ERROR2$：scanf(%d,&n);$ERROR3$：result=fun(n);#include int gcd(int m,int n) int r,t; if(m0) r=m%n; m=n; n=r; return(_); /*$BLANK2$*/ void main() int m,n,result; printf(Pease input two number:n); scanf(%d%d,_); /*$BLANK3$*/ result=gcd(m,n); printf(result=%dn,result);$BLANK1$：n=t$BLANK2$：m$BLANK3$：&m,&n#include void PRINT(long s) FILE *out; printf(s=%ldn,s); if(out=fopen(C:24000108RESULT.DAT,w+)!=NULL) fprintf(out,s=%ld,s); fclose(out);void main()long s=0,t=1,i;for(i=1;i=16;i+) t=t*2; s=s+t; PRINT(s);模拟09#include long fun(int n) if(n=1) /*$ERROR1$*/ return 2; else return fun()+3; /*$ERROR2$*/void main() long m; int n; scanf(%d,n); /*$ERROR3$*/ if(n0) m=fun(n); printf(%ldn,m); $ERROR1$：if(n=1)$ERROR2$：return fun(n-1)+3;$ERROR3$：scanf(%d,&n);#includevoid main()char s80; int i,j;printf(Please input a string:);gets(_); /*$BLANK1$*/for(i=0,j=0;si!=_;i+) /*$BLANK2$*/ if(si != 32) sj+=si; sj=_; /*$BLANK3$*/puts(s);$BLANK1$：s$BLANK2$：0$BLANK3$：0#include void PRINT(double s) FILE *out; printf(s=%.5fn,s); if(out=fopen(C:24000109RESULT.DAT,w+)!=NULL) fprintf(out,s=%.5f,s); fclose(out);void main()double s=0; long i,t=0; for(i=1;i6;i+) t=t*10+8;s=s+1.0/t PRINT(s);172011年省级二级上级模拟复习模拟1二、改错将1到100间的所有素数保存在数组prime中，输出这些素数并计算它们的和（说明：每行输出10个数据，每个数据占4列）#include void main() int i,j,k,sum; int prime50; sum=1; /*$ERROR1$*/ k=0; for(i=2;i100;i+) for(j=2;ji) /*$ERROR2$*/ sum+=i; primek+=i; printf(nPrime list(1-100):); for(i=0;ik;i+) if(i%10=0) printf(n); printf(%4d,&primei); /*$ERROR3$*/ printf(nsum=%dn,sum); $1$:sum=0;$2$:if(i=j)$3$:printf(%4d,primei);三、填空将字符串str中的字符以只写方式写入到新建的文本文件out2011.txt中。#include #include #define N 81void main() int i=0; char ch; char strN=2011 C test!; _ *fp; /*$BLANK1$*/ if(fp=fopen(out2011.txt,_)=NULL) /*$BLANK2$*/ printf(Can not create the file!);exit(0); while(stri) ch=stri; fputc(ch,_);/*$BLANK3$*/ putchar(ch); i+; putchar(n); fclose(fp); BLANK1:FILEBLANK2:wBLANK3:fp四、程序设计求100-999之间的回文素数之和。（1）回文素数是一个素数，且从左向右和从右向左读是相同的，如：101，131，181是回文素数；（2）要求使用循环实现。#include void PRINT(long s) FILE *out; printf(s=%ldn,s); if(out=fopen(C:24000101RESULT.DAT,w+)!=NULL) fprintf(out,s=%ld,s); fclose(out);void main() long s=0; int a,b,i,j; for(i=100;i1000;i+) a=i/100; b=i%10; if(a=b) for(j=2;ji/2) s=s+i; PRINT(s);模拟2二、改错在屏幕上输出以下图形（说明：星号之间没有空格）#include void main() int i,j; i=1; /*$ERROR1$*/ while(i5) j=0; while(j0) j=0; while(j5-i) /*$ERROR2$*/ printf(*); j+; printf(n); i+ /*$ERROR3$*/ $1$:i=0;$2$:while(j=5-i)$3$:i-;三、填空输出以下图形（说明：字符之间没有空格）AAABAAAAABBBAAABBBBBABBBBBBBAABBBAAAAABAAA#include #include void main() int i,j; for(i=-3;i=_;i+) /*$BLANK1$*/ j=1; while(j=abs(i) printf(A);j+; j=1; while(j_7-2*abs(i) /*$BLANK2$*/ printf(B);j+; j=1; while(j=abs(i) printf(A); _; /*$BLANK3$*/ printf(n); BLANK1:3BLANK2:=BLANK3:j+四、程序设计三色球问题若一个口袋中放有12个球，其中有3个红色的，3个白色的和6个黑色的，从中任取8个球，求共有多少种不同的颜色搭配。如：2个白色球和6个黑色球；1个红色球、3个白色球和4个黑色球；#include void PRINT(int n) FILE *out; printf(count=%dn,n); if(out=fopen(C:24000102RESULT.DAT,w+)!=NULL) fprintf(out,count=%dp,n); fclose(out);void main()int n; int a,b,c; for(a=0;a=3;a+) for(b=0;b=3;b+) for(c=0;c=6;c+) if(a+b+c=8) n+; PRINT(n);模拟3二、改错计算9到999之间个位数是9并且能被9整除的所有数之和。#include void main() int i,sum; sum=1; /*$ERROR1$*/ for(i=9;i=999;i+) if(i%9=0) if(i%10=9) sum=i; /*$ERROR2$*/ print(%dn,sum); /*$ERROR3$*/$1$:sum=0;$2$:sum=sum+i;$3$:printf(%dn,sum);三、填空输出所有“水仙花数”，并输出“水仙花数”的个数。所谓“水仙花数”是指一个三位数，其各位数字的立方和等于该数本身。例如153是一个“水仙花数”，因为153=1*1*1+5*5*5+3*3*3。#include void main() int i,a,b,c,n; n=_; /*$BLANK1$*/ for(i=101;i=999;i+) a=i%10; b=i/10%10; c=_; /*$BLANK2$*/ if(a*a*a+b*b*b+c*c*c=i) printf(%6d,i); _; /*$BLANK3$*/ printf(nn=%dn,n);BLANK1:0BLANK2:i/100BLANK3:n+或+n四、程序设计计算1-1000之间能被3或7整除的所有自然数的倒数之和。说明：（1）结果保留5位小数；（2）要求使用循环实现。#include void PRINT(double s) FILE *out; printf(s=%.5fn,s); if(out=fopen(C:24000103RESULT.DAT,w+)!=NULL) fprintf(out,s=%.5f,s); fclose(out);void main() double s=0; int i; for(i=1;i=1000;i+) if(i%3=0|i%7=0) s+=1.0/i; PRINT(s);模拟4二、改错在屏幕上输出以下图形（字符之间没有空格）1234523451345124512351234#include void main() char a6=12345,temp; int i,k; for(i=1;i=5;i+) printf(%cn,a); /*$ERROR1$*/ temp=a0; for(k=0;k4;k+) ak+1=ak; /*$ERROR2$*/ a4=a0; /*$ERROR3$*/ $1$:printf(%sn,a);$2$:ak=ak+1;$3$:a4=temp;三、填空假设银行整存整取存款不同的期限的月息利率分别为：0.315% 期限一年0.330% 期限二年0.345% 期限三年0.375% 期限五年0.420% 期限八年或八年以上输入存钱的本金和整存整取得期限，求到期时能从银行得到的利息与本金的合计。#include #include void main( ) int year; double money,rate,interest,total; printf(Input money and year :n); scanf(%lf%d,&money, &year); if(year1时F(n)=(F(n-1)+10/F(n-1)/2求该数列第5项即F(5)的值。说明：（1）结果保留五位小数；（2）要求使用循环实现。#include void PRINT(double f) FILE *out; printf(F(5)=%.5fn,f); if(out=fopen(C:24000104RESULT.DAT,w+)!=NULL) fprintf(out,F(5)=%.5f,f); fclose(out);void main() double f=5; int i; for(i=2;i=5;i+) f=(f+10/f)/2; PRINT(f);模拟5二、改错把字符串中所有数字字符按以下规律改写（1）0，1，2，3，4，5，6，7，8改写成1，2，3，4，5，6，7，8，9（2）9改成0（3）其他字符保持不变#include #include void main() char s80; int i; getc(s); /*$ERROR1$*/ for(i=0; si!=0; i+) if(si=9) /*$ERROR2$*/ si=0; else if(si=0 & si=8) si=si+1; printf(%cn,s); /*$ERROR3$*/$1$:gets(s);$2$:if(si=9)$3$:printf(%sn,s);三、填空某部门有三位职工，要求输入职工的工资信息，计算每位职工的实发工资和部门实发工资总额（total）（1）描述职工工资信息的数据包括编号（num）、姓名（name）、基本工资（bwage）、奖金（bonus）、保险（ins）和实发工资（rwage）;（2）实发工资=基本工资+奖金-保险（3）职工编号长度小于10位且为整数，姓名不包含空格。#include #define N 3struct employee long num; char name15; long bwage,bonus,ins,rwage;void main() _ workerN;/*$BLANK1$*/ int i; long total; total=_; /*$BLANK2$*/ printf(Please input data:n); printf(nnum name bwage bonus ins:n); for(i=0;iN;i+) scanf(%ld%s, &workeri.num, ); scanf(%ld%ld%ld, &workeri.bwage, &workeri.bonus, &workeri.ins); for(i=0;iN;i+) _=workeri.bwage +workeri.bonus -workeri.ins; /*$BLANK3$*/ total=total+workeri.rwage; printf(output:n ); printf(nnum name real wagen); for(i=0;iN;i+) printf(%-9ld %-14s %-ldn, workeri.num, , workeri.rwage); printf(ntotal wage =%ldn,total);BLANK1: struct employeeBLANK2: 0BLANK3: workeri.rwage