实验三

include <stdio.h>

char score_to_grade(int score);  // 函数声明
int main() {
  int score;
  char grade;
  while(scanf("%d", &score) != EOF) {
    grade = score_to_grade(score);  // 函数调用
    printf("分数: %d, 等级: %c\n\n", score, grade);
 }
  return 0;
}
// 函数定义
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;
}

q1 :功能:分得对应等级。 形参类型:整型 返回值:字符

q2 :ans是单字符,不能用双引号。

include <stdio.h>

int sum_digits(int n); // 函数声明

int main() {
int n;
int ans;

while(printf("Enter n: "), scanf("%d", &n) != EOF) {
ans = sum_digits(n); // 函数调用
printf("n = %d, ans = %d\n\n", n, ans);
}
return 0;
}

// 函数定义
int sum_digits(int n) {
int ans = 0;

while(n != 0) {
ans += n % 10;
n /= 10;
}

return ans;
}
回答1:将所有位加在一起

回答2:可以

include <stdio.h>

int power(int x, int n); // 函数声明
int main() {
int x, n;
int ans;
while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {
ans = power(x, n); // 函数调用
printf("n = %d, ans = %d\n\n", n, ans);
}
return 0;
}
// 函数定义
int power(int x, int n) {
int t;
if(n == 0)
return 1;
else if(n % 2)
return x * power(x, n-1);
t = power(x, n/2);
return t*t;
}
}

B820756E292FCA82435D729A8F30607B

4.#include <stdio.h>
int classify_triangle(int a, int b, int c){
if(a+b>c && a+c>b && b+c>a){
if(ab && bc) return 2;
else if(ab || ac || bc) return 3;
else if(aa+bb
cc || cc+bb==aa || aa+ccb*b) return 4;
else return 1;
}
else return 0;
}
int main()
{
int a, b, c;
while(scanf("%d%d%d", &a, &b, &c) != EOF){
if(classify_triangle(a,b,c)
0) printf("不能构成三角形\n");
if(classify_triangle(a,b,c)1) printf("普通三角形\n");
if(classify_triangle(a,b,c)
2) printf("等边三角形\n");
if(classify_triangle(a,b,c)3) printf("等腰三角形\n");
if(classify_triangle(a,b,c)
4) printf("直角三角形\n");
}
return 0;
}

联想截图_20260420114112

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 || m > n) return 0;
if (m == 0 || m == n) return 1;
if (m > n - m) m = n - m;
int result = 1;
for (int i = 1; i <= m; i++) {
result = result * (n - m + i) / i;
}
return result;
}

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||m>n) return 0;
if(m0||mn) return 1;
return func(n-1,m)+func(n-1,m-1);
}

1

include <stdio.h>

int gcd(int a, int b, int c);

int main() {
int a, b, c;
int ans;

while (scanf("%d%d%d", &a, &b, &c) != EOF) {
ans = gcd(a, b, c);
printf("最大公约数: %d\n\n", ans);
}

return 0;
}
int gcd(int a,int b,int c){
int min=a,i;
if(b<min) min=b;
if(c<min) min=c;
for (i=min;i>=1;i--)
if (a%i0&&b%i0&&c%i==0)
return i;
return 1;
}
2

7.#include <stdio.h>

void print_charman(int n);

int main()
{
int n;
printf("Enter n: ");
scanf("%d", &n);
print_charman(n);
return 0;
}
void print_charman(int n)
{
int i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<i;j++)
printf("\t");
int count=2*(n-i)-1;
for (k=0;k<count;k++)
printf(" 0\t");
printf("\n");
for(j=0;j<i;j++)
printf("\t");
for(k=0;k<count;k++)
printf("\t");
printf("\n");
for(j=0;j<i;j++)
printf("\t");
for(k=0;k<count;k++)
printf("I I\t");
printf("\n\n");
}
}

11

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posted @ 2026-04-20 11:45  xialkQwQ  阅读(11)  评论(0)    收藏  举报