湖工大C语言题库.docx

1.求A数组中奇数的和。#include #include void main() int arr10=11,4,2,7,3,12,5,34,5,9,i; int s=0; /*found*/ for ( i=0; i10; i+) /*found*/ if (arri% 2 = 1) s = s + arri; printf(The result is: %dn, s); 2.求一维数组中素数的和2,3,5,7,8,9，10，11,12,13,15,17.#include #include #include void main() int a12=2,3,5,7,8,9,10,11,12,13,15,17,i,j,k,s; /*found*/ s=0; for ( i=0; i12; i+) k=(int)sqrt(ai); for ( j=2;j k ) /*found*/ s=s+ai; printf(Sum is: %dn,s); 3. 求数组中 所有奇数的平均值#include#includedoubleaverage(intarr,intn)intk=0,i;doubles;s=0;for(i=0;in;i+)/*found*/if(arri%2=1)s=s+arri;k+;return(s/k);voidmain()inta12=10,4,2,7,3,12,5,34,5,9,21,18;doubles;/*found*/s=average(a,12);printf(Theresultis:%.2fn,s);4.求【m n】中既不能被7整除也不能被5整除的和#include stdio.h int fun(int m,int n) /*begin*/int i,sum=0;for(i=m;i=n;i+)if(i%7)!=0)&(i%5)!=0)sum=sum+i;return sum;/*end*/ void main() void NONO( );/函数声明 int s; int m,n; printf(Enter m n:n); scanf(%d %d,&m,&n); s=fun(m,n); printf(s=%dn,s); NONO( ); void NONO( ) FILE *fr,*fw; int i,x,y; fr=fopen(D:exam03910431PROGIN.DAT,r); fw=fopen(D:exam03910431PROGOUT.DAT,w); for(i=1;i=5;i+) fscanf(fr,%d,%d,&x,&y); fprintf(fw,The Result is :%d*n,fun(x,y); fclose(fr); fclose(fw); 56。输入一个百分制成绩，打印出五级计分成绩，考试在90或90以上为优秀，80-89为良好，70-79为中等，60-69为及格，60以下为不及格。#include #include #include #include void main() int score,t; printf(Please enter a score:); do scanf(%d,&score); while(score100); t=score/10; /*found*/ switch(t) case 10: case 9:printf(优秀!n);break; case 8:printf(良好!n);break; case 7:printf(中等!n);break; case 6:printf(及格!n);break; /*found*/ default:printf(不及格!n); 7在C程序中打印出1到1000中满足其个位数的立方等于其本身的所有整数#include main() int i,g; for(i=1;i1000;i+) /*found*/ g=i%10; /*found*/ if(g*g*g=i) printf(%4d,i); printf(n); 8求一维数组A中为奇数的元数的平均值元数为10，4，2，7，3，12，5，34，5，9，21，18，输出为8.3#include #include double average( int arr , int n ) int k=0,i; double s; s = 0; for ( i=0 ;in; i+) /*found*/ if (arri % 2 = 1) s = s + arri; k+; return (s/k) ; void main() int a12=10,4,2,7,3,12,5,34,5,9,21,18; double s; /*found*/ s = average(a,12); printf(The result is: %.2fn, s); 9. #include#defineN3intfun(intaNN)intmin;/*begin*/inti,j;min=a00;for(i=0;iN;i+)for(j=0;jN;j+)if(aijmin)min=aij;/*end*/returnmin;voidmain()voidNONO();/函数声明intaNN=-1,2,3,4,9,5,7,8,6;inti,j,min;printf(arrayis:n);for(i=0;iN;i+)for(j=0;jN;j+)printf(%5d,aij);printf(n);min=fun(a);printf(Resultis:%dn,min);NONO();voidNONO()FILE*fr,*fw;inti,j,k,m,aNN;fr=fopen(D:exam03910411PROGIN.DAT,r);fw=fopen(D:exam03910411PROGOUT.DAT,w);for(i=1;i=5;i+)for(j=0;jN;j+)for(k=0;kN;k+)fscanf(fr,%d,&ajk);fprintf(fw,%4d,ajk);fprintf(fw,Min=%dn,fun(a);fclose(fr);fclose(fw);10.#include stdio.h double fun( int n ) double c; /*begin*/ c=0.0 ; int i,t=1; for(i=1;i=n;i+) c=c+t*1.0/(2*i-1);t=-t; return c;/*end*/ void main() void NONO( );/函数声明 double sum; int n; printf(Enter n:n); scanf(%d,&n); sum=fun(n); printf(sum=%.4fn,sum); NONO( ); void NONO( ) FILE *fr,*fw; int i, x; fr=fopen(D: exam03910304PROGIN.DAT,r); fw=fopen(D: exam03910304PROGOUT.DAT,w); for(i=1;i=5;i+) fscanf(fr,%d,&x); fprintf(fw,s=%.4fn,fun(x);fclose(fr); fclose(fw);11.i=m b=b+(i+1)12.将数组a的5个元素对折后再输出，比如：a5=1,2,3,4,5,对折后输出1,2,3,2,1 #define M 5 void main( ) int i,a100,t; /clrscr(); for(i=0;iM;i+) /*found*/ _(1)_ /答案：scanf(%d,&ai); /*found*/ for(i=0;_(2)_;i+) /答案：i2/M aM-i-1=ai; for(i=0;iM;i+) printf(%5d,ai); printf(n); 13.求数组中值为偶数的元数和#include #include sum ( int arr ,int n ) int i,s; s = 0; for ( i=0; i=x) s =s+t; n=n+2.0; k=-k; t=k/n; return s ; /程序编写部分 /*end*/ void main() double x,sum; printf(Enter x:n); scanf(%lf,&x); sum=fun(x); printf(sum=%.4fn,sum); NONO( ); NONO( ) FILE *fr,*fw; int i; double x; fr=fopen(D:exam04500129PROGIN.DAT,r); fw=fopen(D:exam04500129PROGOUT.DAT,w); for(i=1;i=5;i+) fscanf(fr,%lf,&x); fprintf(fw,s=%.4fn,fun(x); fclose(fr); fclose(fw); 15. 输出fabonacii数列1,1,2,3,5,8,11，。前20项，每5个数换一行#include stdio.h void main() long int f20=1,1; int i; /clrscr(); /*found*/ _(1)_;/填：printf(%8ld%8ld,f0,f1); for(i=2;i20;i+) fi=fi-1+fi-2; /*found*/ if(_(2)_) printf(n);/填：i%5=0 printf(%8ld,fi); printf(n); 16.公倍数#includestdio.h#includemath.h#includeconio.h#includestdlib.hvoidmain()intm=76,n=40,t,i;/*begin*/i=m%n;t=m*n;while(i!=0)m=n;n=i;i=m%n;/辗转相除法t=t/n;/最小公倍数等于两数乘积/最大公约数/*end*/printf(%d%dn,n,t);NONO(m,n,t);NONO(intm,intn,in