
思路整体向下向右平均2个单位,这样就不用考虑越界问题了。
最大值20,所以开数组时,下标开到25就行。
由于产生的数可能非常大,dp数组的值要用 long long.
统一坐标。
c
#include <stdio.h>
#define N 24
typedef struct post {
int x; //水平
int y; //垂直
}post;
int arr[N][N] = {0};
post a, b, c;
long long dp[N][N]; //dp[i][j] 指的是到i j位置时,经过的数量
int main()
{
int tmp = scanf("%d%d%d%d", &b.x, &b.y, &c.x, &c.y);
// 整体向下,向右平移2个单位
a.x = 2;
a.y = 2;
b.x += 2;
b.y += 2;
c.x += 2;
c.y += 2;
//根据马的位置,把相关位置堵死
arr[c.x][c.y] = 1;
arr[c.x - 2][c.y - 1] = 1;
arr[c.x + 2][c.y - 1] = 1;
arr[c.x - 2][c.y + 1] = 1;
arr[c.x + 2][c.y + 1] = 1;
arr[c.x - 1][c.y - 2] = 1;
arr[c.x + 1][c.y - 2] = 1;
arr[c.x - 1][c.y + 2] = 1;
arr[c.x + 1][c.y + 2] = 1;
//dp初始化
dp[a.x][a.y] = 1;
for (int i = 2; i <= b.x; i++) {
for (int j = 2; j <= b.y; j++) {
if (i == 2 && j == 2) {
continue;
}
if (arr[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
else {
dp[i][j] = 0;
}
//printf("%lld ", dp[i][j]);
}
//printf("\n");
}
printf("%lld", dp[b.x][b.y]);
return 0;
}
c
#include <stdio.h>
#include <string.h>
#define N 24
typedef struct post {
int x; //水平
int y; //垂直
} post;
int arr[N][N];
post a, b, c;
long long dp[N][N]; //dp[i][j] 指的是到i j位置时,经过的数量
int main() {
// 输入数据
scanf("%d%d%d%d", &b.x, &b.y, &c.x, &c.y);
// 整体向下,向右平移2个单位
a.x = 2;
a.y = 2;
b.x += 2;
b.y += 2;
c.x += 2;
c.y += 2;
// 初始化arr和dp数组为0
memset(arr, 0, sizeof(arr));
memset(dp, 0, sizeof(dp));
// 标记马的控制点
arr[c.x][c.y] = 1;
int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};
for (int k = 0; k < 8; k++) {
int nx = c.x + dx[k];
int ny = c.y + dy[k];
if (nx >= 0 && nx < N && ny >= 0 && ny < N) {
arr[nx][ny] = 1;
}
}
// dp初始化
dp[a.x][a.y] = 1;
for (int i = a.x; i <= b.x; i++) {
for (int j = a.y; j <= b.y; j++) {
if (i == a.x && j == a.y) {
continue;
}
if (arr[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
} else {
dp[i][j] = 0;
}
}
}
printf("%lld\n", dp[b.x][b.y]);
return 0;
}