10 Wrong Answer
# | 状态 分数 | 耗时 | 内存占用 |
---|---|---|---|
#1 | Wrong Answer 0 | 2ms | 764 KiB |
#2 | Wrong Answer 0 | 1ms | 288 KiB |
#3 | Wrong Answer 0 | 1ms | 324 KiB |
#4 | Wrong Answer 0 | 2ms | 516 KiB |
#5 | Wrong Answer 0 | 1ms | 288 KiB |
#6 | Wrong Answer 0 | 2ms | 320 KiB |
#7 | Wrong Answer 0 | 1ms | 288 KiB |
#8 | Wrong Answer 0 | 1ms | 288 KiB |
#9 | Wrong Answer 0 | 1ms | 324 KiB |
#10 | Accepted 10 | 1ms | 324 KiB |
代码
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
int matrix[n][n];
int num = 1;
int left = 0, right = n - 1, top = 0, bottom = n - 1;
while (left <= right && top <= bottom) {
// 从左到右填充上侧
for (int i = left; i <= right; i++) {
matrix[top][i]=num++;
}
top++;
// 从上到下填充右侧
for (int i = top; i <= bottom; i++) {
matrix[i][right]=num++;
}
right--;
if (top <= bottom) {
// 从右到左填充下侧
for (int i = right; i >= left; i--) {
matrix[bottom][i]=num++;
}
bottom--;
}
if (left <= right) {
// 从下到上填充左侧
for (int i = bottom; i >= top; i--) {
matrix[i][left]=num++;
}
left++;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// 按照场宽为4输出
printf("%4d", matrix[i][j]);
}
printf("\n");
}
return 0;
}
信息
- 递交者
- 题目
- P369 练63.3 迂回方阵
- 语言
- C
- 代码长度
- 1 KiB
- 递交时间
- 2024-11-9 16:42:51
- 评测时间
- 2024-11-9 16:42:52
- 分数
- 10
- 总耗时
- 13ms
- 峰值时间
- 2ms
- 峰值内存
- 764 KiB