实验3
task.1
#include <stdio.h> char score_to_grade(int score); int main() { int score; char grade; while(scanf("%d", &score) != EOF) { grade = score_to_grade(score); printf("分数: %d, 等级: %c\n\n", score, grade); } return 0; } char score_to_grade(int score) { char ans; switch(score/10) { case 10: case 9: ans = 'A'; break; case 8: ans = 'B'; break; case 7: ans = 'C'; break; case 6: ans = 'D'; break; default: ans = 'E'; } return ans; }
#include <stdio.h> char score_to_grade(int score); int main() { int score; char grade; while(scanf("%d", &score) != EOF) { grade = score_to_grade(score); printf("分数: %d, 等级: %c\n\n", score, grade); } return 0; } char score_to_grade(int score) { char ans; switch(score/10) { case 10: case 9: ans = 'A'; case 8: ans = 'B'; case 7: ans = 'C'; case 6: ans = 'D'; default: ans = 'E'; } return ans; }
问题1:将分数分为“A”，“B”，“C”，“D”,“E”五个等级。 整数型 字符型
问题2:有问题，无论输入数字为什么，均会输出“E”。
task.2
#include <stdio.h> int sum_digits(int n); int main() { int n; int ans; while(printf("Enter n: "), scanf("%d", &n) != EOF) { ans = sum_digits(n); printf("n = %d, ans = %d\n\n", n, ans); } return 0; } int sum_digits(int n) { int ans = 0; while(n != 0) { ans += n % 10; n /= 10; } return ans; }
#include <stdio.h> int sum_digits(int n); int main() { int n; int ans; while(printf("Enter n: "), scanf("%d", &n) != EOF) { ans = sum_digits(n); printf("n = %d, ans = %d\n\n", n, ans); } return 0; } int sum_digits(int n) { if(n<10) return n; return sum_digits(n/10) + n%10; }
问题1：将输入数的各位相加输出和
问题2:效果相同，算法1直接通过递归算出各位数字之和；算法2先判断n与10的大小，小于10直接输出，大于10同算法1.
task.3
#include <stdio.h> int power(int x, int n); int main() { int x, n; int ans; while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) { ans = power(x, n); printf("n = %d, ans = %d\n\n", n, ans); } return 0; } int power(int x, int n) { int t; if(n == 0) return 1; else if(n % 2) return x * power(x, n-1); else { t = power(x, n/2); return t*t; } }
问题1：计算x的n次方
问题2：
task.4
#include <stdio.h> #include <math.h> #include <stdlib.h> int is_prime(int n); int main() { int i, count=0; for(i = 2;i <=100;i++) { if(is_prime(i)&&is_prime(i+2)) printf("%d,%d\n",i, i+2); count++; } printf("100以内的孪生素数共有%d",count); return 0; } int is_prime(int n) { int i; for(i = 2;i<=sqrt(n);i++) if(n%i==0) return 0; return 1; }
task.5
#include <stdio.h> #include <stdlib.h> void hanoi(unsigned int n,char from,char temp,char to); void moveplate(unsigned int n,char from,char to); int count=0; int main() { unsigned int n; while(scanf("%u",&n)!=EOF) { hanoi(n,'A','B','C'); printf("一共移动了%d次\n\n",count); count=0; } system("pause"); return 0; } void hanoi(unsigned int n,char from,char temp,char to) { if(n == 1) { moveplate(n,from,to); count++; } else { hanoi(n-1,from,to,temp); moveplate(n,from,to); count++; hanoi(n-1,temp,from,to); } } void moveplate(unsigned int n,char from,char to) { printf("%u:%c-->%c\n\n",n,from,to); }
task.6
递归：
#include <stdio.h> int func(int n, int m); int main() { int n, m; int ans; while(scanf("%d%d", &n, &m) != EOF) { ans = func(n, m); printf("n = %d, m = %d, ans = %d\n\n", n, m, ans); } return 0; } int func(int n, int m) { if(m = 0 || n == m) return 1; else if(n < m) return 0; else return func(n-1,m)+func(n-1,m-1); }
迭代：
#include <stdio.h> int func(int n, int m); int main() { int n, m; int ans; while(scanf("%d%d", &n, &m) != EOF) { ans = func(n, m); printf("n = %d, m = %d, ans = %d\n\n", n, m, ans); } return 0; } int func(int n, int m) { int i, j,up, down; for(i=n;i>=(m-1);i--) { up=up*i; } for(j=m;j>=1;j--) { down=down*j; } return up/down; }
task.7
#include <stdio.h> #include <stdlib.h> void print_charman(int n); int main() { int n; printf("Enter n: "); scanf("%d", &n); print_charman(n); // 函数调用 return 0; } void print_charman(int n) { int m = ((n*2)-1) ,count=0,i,x,b,a; while(m!=-1){ for(x=0;x<count;x++){ printf("\t"); } for(i=0; i<m ;i++){ printf(" O\t"); } printf("\n"); for(x=0;x<count;x++){ printf("\t"); } for(a=0; a<m ;a++){ printf("<H>\t"); } printf("\n"); for(x=0;x<count;x++){ printf("\t"); } for(b=0; b<m ;b++){ printf("I I\t"); } printf("\n"); count++; m=m-2; } }
