PTA程序设计答案.docx

1、PTA程序设计答案PTA所有答案工业gc1601学习小组一、判断题二、选择题 三、函数题4-3 求m到n之和 (10分)答案：int sum(int m, int n) int s=0,i; for(i=m;i=n;i+) s+=i; return s;4-34 简单输出整数 (10分)答案：void PrintN ( int N ) int i; for(i=1;i=N;i+) printf(%dn,i);4-36 简单求和 (10分)答案：int Sum ( int List, int N ) int i,s=0; for(i=0;iN;i+) s+=Listi; return s;4-4

2、0 简单阶乘计算 (10分)答案：int Factorial( const int N ) int i, s=1; if( N 0 ) return 0; else if( N = 0) return 1; else for(i=1; i= b) max = a; else max = b; return max;4-44 数字金字塔 (15分)答案：void pyramid(int n) int i, j, space; for(i=1; i = n; i+) space = n - i; for(j=0; j space; j+) printf( ); for(j=0; j i; j+)

3、printf(%-2d, i); printf(n); 4-45 判断奇偶性 (10分)答案：int even( int n ) if( n%2=0 ) return 1; else return 0;4-46 使用函数求奇数和 (15分)答案：int even( int n ) if( n%2=0 ) return 1; else return 0;int OddSum( int List, int N ) int i; int oddsum = 0; for(i=0; iN; i+) if(even(Listi)=0) oddsum += Listi; return oddsum;4-47

4、 使用函数计算两点间的距离 (10分)答案：double dist(double x1, double y1, double x2, double y2) double d; d = sqrt(x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2); return d;4-48 使用函数求素数和 (20分)答案：int prime( int p ) int i,j,count =0; for(j=2;j0&p!=1) return 1; else return 0;int PrimeSum( int m, int n ) int i,sum=0; for(i=m;i

5、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; if (powerSum = number) flag = 1; else flag = 0; return flag;void PrintN(int

6、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= a&si = A&si = 0&si = 9) n3+; else n4+; printf(letter = %d, blank = %d, digit = %d, other = %d,n1,n2,n3,n4);4-7 使用函数求特殊a串数列和 (20分)答案：int fn( int a, int

7、n ) int x,p=0,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; 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)答案：int factorsum( int number ) int i,s=1; for(i=1; i1 & number%i = 0) s+=i; return s;void PrintPN( int m,

8、 int n ) int k,t,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数的逆序数 (20分)答案：int reverse( int number ) int m=0; if(number 0) m = m*10 + number%10; number=nu

9、mber/10; return m;4-50 移动字母 (10分)答案：void Shift( char s ) int i,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;4-51 在数组中查找指定元素 (15分)答案：int search(int list, int 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;

10、 i= A&si = a&si = 0&si = 9) n4+; else n5+; printf(%d %d %d %d %d,n1,n2,n3,n4,n5);4-26 使用递归函数计算1到n之和 (10分)答案：int sum( int n ) int i,sum=0; if(n=0) return 0; else for(i=1; i=n; i+) sum += i; return sum; 4-28 递归求阶乘和 (15分)答案：double fact( int n ) int i; double m=1; if(n=0) return m; else for(i=1; i0) for

11、(j=1; j=n; j+) sum += fact(j); return sum;4-29 递归实现指数函数 (15分)答案：double calc_pow( double x, int n ) int i; double m=1; for(i=1; i=n; i+) m = m*x; return m;4-32 递归求Fabonacci数列 (10分)答案：int f( int n ) int d; if(n=0) d=0; else if(n=1) d=1; else d=f(n-2)+f(n-1); return d;4-33 十进制转换二进制 (15分)答案：void dectobi

12、n( int n ) int t; if(n=0|n=1) printf(%d,n%2); else t=n; dectobin(n=n/2); printf(%d,t%2); 4-8 递归实现顺序输出整数 (15分)答案：void printdigits(int n) if (n 0) s += number%10; number /= 10; if(s = 5) return 1; else return 0;void count_sum( int a, int b ) int i, count=0, count_sum=0; for(i=a; i=b; i+) if( is(i) ) c

13、ount+; count_sum += i; printf(count = %d, sum = %d, count, count_sum);4-10 简单实现x的n次方 (10分)答案：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)答案：double fact( int n ) int i; double s=1; for(i=1;i=n; i+) s = s * i; return s;4-14 使用函数求最大公约数 (

14、10分)答案：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)答案：void sort( int a, int n ) int i,j,k,t; for(i=0;in-1;i+) for(j=i+1;jn;j+) if(ajai) t=ai;ai=aj;aj=t; 4-16 函数实现字符串逆序 (15分)答案：void f(char *p) int i,n=0; char t; for(i=0; pi != 0; i+) n+; for (i=0

15、; i 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)答案：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)答案

16、：double dist( double h, double p ) double d=h; while(p!=0) h=p*h; if(hTOL) break; d=d+2*h; return d;四、编程题单词长度 (20分)答案：#include #include int main() char a1000; int i, n = 0, k, t = 0; for (i = 0; i+) scanf(%c, &ai); if (ai = .) break; k = i; for (i = i - 1; i = 0; i-) if (ai = ) t+; else break; for (

17、i = 0; i = k; i+) if (ai = .) if (n != 0) printf(%d, n); printf(n); break; if (ai = ) if (n != 0) printf(%d, n); n = 0; if (i k - t) printf( ); else n+; return 0;掉入陷阱的数字 (20分)答案：#include int main() int n1000, s1000, i, t; scanf(%d, &n0); for (i = 1; i 0) si - 1 += t % 10; t /= 10; ni = 3 * si - 1 +

18、1; printf(%d:%dn, i, ni); if (ni = ni - 1)break; return 0;九宫格输入法 (20分)答案：#include #includeint main() char *key10 = 0 , 1,.?!, 2ABC, 3DEF, 4GHI, 5JKL, 6MNO, 7PQRS, 8TUV, 9WXYZ ; int length10; for (int i = 0; i10; i+) lengthi = strlen(keyi); char ch; char p; int count = 0; do ch = getchar(); if (ch != &ch != n) p = ch; count+; else putchar(keyp - 0(count - 1) % lengthp - 0); count = 0; while (ch != n); return 0;算术入门之加减乘除 (20分)答案：#include main() int a, b, r1, r2, r3, r41, c; floa