PTA函数答案.docx

1、PTA函数答案三、程序填空题3-32程序阅读：下列程序将输出 (5分)。# include int s;int f(int m) static int k=0; for(; k=m; k+) s+; return s;int main(void ) int s=1; s=f(2)+f(2); printf(%d#%d#, s, f(20); return 0;四、函数4-3求m到n之和(10分)int sum(int m, int n) int i; int s=0; for(i=m; i=n; i+) s= s + i; return (s);4-34简单输出整数(10分)void Prin

2、tN ( int N ) int i; for(i=1; i=N; i+) printf(%dn, i);4-36简单求和(10分)int Sum ( int List, int N ) int k; int sum=0; for(k=0; kN; k+) sum += Listk; return sum;4-40简单阶乘计算(10分)int Factorial( const int N ) int i; int s=1; if( N 0 ) return 0; else if( N = 0) return 1; else for(i=1; i=N; i+) s*=i; return s;4-

3、41统计个位数字(15分)int Count_Digit(const int N, const int D) int x; int count=0; int temp=0; x=N; if(x0); return count; 4-43找两个数中最大者(10分)int max( int a, int b ) int max; if(a = b) max = a; else max = b; return max;4-44数字金字塔(15分)void pyramid(int n) int i, j, space; for(i=1; i = n; i+) space = n - i; for(j=

4、0; j space; j+) putchar( ); for(j=0; j i; j+) printf(%-2d, i); putchar(n); 4-45判断奇偶性(10分)int even( int n ) if( n0 ) n = -n; if( n%2=0 ) return 1; else return 0;4-46使用函数求奇数和(15分)int even( int n ) if( n0 ) n = -n; if( n%2=0 ) return 1; else return 0;int OddSum( int List, int N ) int k; int oddsum = 0;

5、 for(k=0; kN; k+) if(even(Listk)=0) oddsum += Listk; return oddsum;4-47使用函数计算两点间的距离(10分)double dist( double x1, double y1, double x2, double y2 ) double Dist; Dist = sqrt(x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ); return Dist;4-48使用函数求素数和(20分)int prime( int p ) int i,j; int count =0; for(j=2;j0&p!=1) retur

6、n 1; else return 0; int PrimeSum( int m, int n ) int i,sum=0; for(i=m;i=n;i+) if(prime(i)=1) sum=sum+i; return sum; 4-19使用函数统计指定数字的个数(15分)int CountDigit( int number, int digit ) int count,temp,x; x=number; if(x0); return count; 4-20使用函数输出水仙花数(20分)int narcissistic(int number) int i, j, digit, power,

7、powerSum, integer, count, flag; count = 0; integer = number; while (integer 0) integer = integer / 10; count+; powerSum = 0; integer = number; for (i = 0; i count; i+) digit = integer % 10; integer = integer / 10; power = 1; for (j = 0; j count; j+) power = power*digit; powerSum = powerSum + power;

8、if (powerSum = number) flag = 1; else flag = 0; return flag;void PrintN(int m, int n) int i; for (i = m + 1; i e) count+; if(count%2!=0) cj=1; for(j=1; j=i; j+) cj=cj*j; d=pow(x,i)/cj; sum=sum+d; i+=2; else cj=1; for(j=1; j=i; j+) cj=cj*j; d=pow(x,i)/cj; sum=sum-d; i+=2; return sum;4-49分类统计字符个数(15分)

9、void f( long int x, char *p ) char q; static int i=0,j=0; if(x0) x=-x; *p=-; p+; q=x%16; if(q0) f(x, p); *(p+(i-j)=q; j-; 4-7使用函数求特殊a串数列和(20分)int fn( int a, int n ) int x; int p=0; int i; for(i=0; in; i+) x= a * pow(10,i); p = p + x; return p;int SumA( int a, int n ) int k; int sum=0; for(k=0; kn+1;

10、 k+) sum = sum + fn( a , k ); return sum;4-22使用函数输出指定范围内的完数(20分)void PrintPN( int m, int n ) int k; int t; int found = 0; for(k=m; k=n; k+) if (factorsum( k ) = k) found = 1; printf(%d = 1,k); for(t=1; t1 & k%t = 0) printf( + %d, t); printf(n); if (found = 0) printf(No perfect number); 4-25使用函数输出一个整

11、数的逆序数(20分)int reverse( int number ) int m=0; if(number 0) m = m*10 + number%10; number=number/10; return m; 4-50移动字母(10分)void Shift( char s ) int i; int n; char a3; n = strlen(s); for(i=0; i3 ) for(i=3; in; i+) si-3 = si; for(i=0; i3; i+) sn-3+i=ai; return s;4-51在数组中查找指定元素(15分)int search(int list, i

12、nt n, int x) int i, result; for (i=0; i=n) m=m%n; if(m!=0) for(i=n-m; i=0; i-) ai+m=ai; for(i=0; i=A & si=a & si=0 & si=9) digit+; else other+; printf(%d %d %d %d %d,bletter,sletter,blank,digit,other);4-26使用递归函数计算1到n之和(10分)int sum( int n ) int i; int Sum=0; if(n=0) return 0; else for(i=1; i=n; i+) S

13、um += i; return Sum;4-28递归求阶乘和(15分)double fact( int n ) int i; double m=1; if(n=0) return m; else for(i=1; i0) for(j=1; j=n; j+) sum += fact(j); return sum;4-29递归实现指数函数(15分)double calc_pow( double x, int n ) int i; double m=1; for(i=1; i9) N+; m=m/10; for(i=N-1; i=0; i-) shang = n * pow(10,-i); yush

14、u = n - shang * pow(10 , i); n = yushu; printf(%dn, shang); 4-9统计各位数字之和是5的数(20分)int is( int number ) int s=0; int yushu = number; while(number 0) yushu = number % 10; s = yushu + s; number = number / 10; if(s = 5) return 1; else return 0;void count_sum( int a, int b ) int i; int count=0; int count_S

15、um=0; for(i=a; i=b; i+) if( is(i) ) count+; count_Sum = count_Sum + i; printf(count = %d, sum = %d, count, count_Sum);4-10简单实现x的n次方(10分)double mypow( double x, int n ) int i; double s=1; for(i=0; in; i+) s = s*x; return s;4-13使用函数求1到10的阶乘和(10分)double fact( int n ) int k; int s=1; for(k=1;k=n; k+) s

16、= s * k; return s;4-14使用函数求最大公约数(10分)int gcd( int x, int y ) int i,t; if(x=1;i-) if(x%i=0 & y%i=0) break; return i;4-15使用函数的选择法排序(25分)void sort( int a, int n ) int temp,k,min,i; for(k=0;kn-1;k+) min=k; for(i=k+1;in;i+) if(ai p) temp = *p; *p = *end; *end = temp; -end; +p; int my_strlen(char *p) char

17、 *q = p; while(*p != 0) p+; return p-q; 4-17递归计算P函数(15分)double P( int n, double x ) double z; if(n = 0) z = 1; else if(n = 1) z=x; else if(n 1) z = (2*n-1)*P(n-1,x) - (n-1)*P(n-2,x)/n; return z;4-1计算存款利息-genlib(10分)int main(void) int money, year; double interest, rate; scanf(%d, &money); scanf(%d, &year); scanf(%lf, &rate); interest = money * pow(1+rate,year) - money; printf(interest = %.2fn, interest); return 0;4-2弹球距离(15分)double dist( double h, double p ) double d=h; while(p!=0) h=p*h; if(hTOL) break; d=d+2*h; return (d);