int a=0; int c=0; do{ –-c; a=a-1; }while(a>0); System.out.println(a+" "+c);
时间: 2024-05-05 22:15:45 浏览: 15
As an AI language model, I can see that the given code will result in a syntax error because of the missing "+" or "-" operator between "a" and the string to print. Here's the corrected code with the assumption that the missing operator is "-":
```
int a = 0;
int c = 0;
do {
--c;
a = a - 1;
} while (a > 0);
System.out.println(a + " " + c);
```
This code will decrement the value of "c" by 1 in each iteration and "a" by 1 until "a" becomes less than or equal to 0. Finally, it will print the value of "a" and "c" separated by a space. Since the initial value of "a" is 0, the loop will not execute even once, resulting in the output "0 -1".
相关问题
int a=1,b=9;do{ b=b-a; a++;}while(b>0);
执行完这段代码后,变量a的值为4,变量b的值为0。
代码的执行过程如下:
1. 变量a被初始化为1,变量b被初始化为9。
2. 进入do-while循环,先执行循环体内的语句。b被赋值为b-a,即b=9-1=8;a被增加1,即a=2。
3. 执行完循环体内的语句后,检查循环条件b>0是否满足,由于b仍然大于0,因此继续执行循环。
4. 再次执行循环体内的语句。b被赋值为b-a,即b=8-2=6;a被增加1,即a=3。
5. 继续检查循环条件b>0是否满足,由于b仍然大于0,因此继续执行循环。
6. 第三次执行循环体内的语句。b被赋值为b-a,即b=6-3=3;a被增加1,即a=4。
7. 继续检查循环条件b>0是否满足,由于b仍然大于0,因此继续执行循环。
8. 第四次执行循环体内的语句。b被赋值为b-a,即b=3-4=-1;a被增加1,即a=5。
9. 再次检查循环条件b>0是否满足,由于b小于等于0,因此跳出循环。
因此,执行完这段代码后,变量a的值为4,变量b的值为0。
#include<stdio.h> #include<stdlib.h> struct SL { int data; struct SL* pre; struct SL* next; }; typedef struct SL* SL; SL buynode(int a) { SL newnode = (SL)malloc(sizeof(struct SL)); newnode->data = a; newnode->next = NULL; newnode->pre = NULL; return newnode; } void Append(SL p, int n) { SL last = p; for (int i = 2; i <= n; i++) { SL newnode = buynode(i); newnode->pre = last; last->next = newnode; newnode->next = p; p->pre = newnode; last = newnode; } } void print1(SL p, int m) { SL pos = p->next; while (pos != NULL) { printf("%4d", pos->data); pos = pos->next; } pos = p->next; while (pos->next != pos) { for (int i = 1; i < m - 1; i++) { pos = pos->next; } SL temp = pos->next; pos->next = temp->next; temp->next->pre = pos; printf("%4d", temp->data); free(temp); pos = pos->next; } printf("%4d\n", pos->data); } void print(SL p, int m, int k,int n) { SL pos = p; int* out_order = (int*)malloc(sizeof(int)*n); int index = 0; while (pos->next != pos) { for (int i = 1; i < m - 1; i++) { pos = pos->next; } SL temp = pos->next; pos->next = temp->next; temp->next->pre = pos; out_order[index++] = temp->data; free(temp); pos = pos->next; } out_order[index++] = pos->data; pos = buynode(out_order[0]); for (int j=1;j<n;j++) { Append(pos, out_order[j]); } print1(pos,k); free(out_order); } int main() { int n, m, k; scanf("%d", &n); scanf("%d", &m); scanf("%d", &k); SL yuese = buynode(1); Append(yuese,n); print(yuese,m,k,n); return 0; }检查代码问题并修改
代码的主要问题在于print函数中没有对空间进行释放,同时在最后一次遍历链表时,应该使用do-while循环,保证最后一个节点也被遍历到。另外,print1函数中的while循环可以和第二个while循环合并成一个do-while循环来简化代码。
修改后的代码如下:
```
#include <stdio.h>
#include <stdlib.h>
struct SL {
int data;
struct SL* pre;
struct SL* next;
};
typedef struct SL* SL;
SL buynode(int a) {
SL newnode = (SL)malloc(sizeof(struct SL));
newnode->data = a;
newnode->next = NULL;
newnode->pre = NULL;
return newnode;
}
void Append(SL p, int n) {
SL last = p;
for (int i = 2; i <= n; i++) {
SL newnode = buynode(i);
newnode->pre = last;
last->next = newnode;
newnode->next = p;
p->pre = newnode;
last = newnode;
}
}
void print1(SL p, int m) {
SL pos = p->next;
do {
for (int i = 1; i < m - 1; i++) {
pos = pos->next;
}
SL temp = pos->next;
pos->next = temp->next;
temp->next->pre = pos;
printf("%4d", temp->data);
free(temp);
pos = pos->next;
} while (pos != p->next);
printf("%4d\n", pos->data);
}
void print(SL p, int m, int k, int n) {
SL pos = p;
int* out_order = (int*)malloc(sizeof(int) * n);
int index = 0;
do {
for (int i = 1; i < m - 1; i++) {
pos = pos->next;
}
SL temp = pos->next;
pos->next = temp->next;
temp->next->pre = pos;
out_order[index++] = temp->data;
free(temp);
pos = pos->next;
} while (pos != p);
out_order[index++] = pos->data;
pos = buynode(out_order[0]);
Append(pos, n - 1);
print1(pos, k);
free(out_order);
}
int main() {
int n, m, k;
scanf("%d", &n);
scanf("%d", &m);
scanf("%d", &k);
SL yuese = buynode(1);
Append(yuese, n);
print(yuese, m, k, n);
return 0;
}
```