P1002 [NOIP 2002 普及组] 过河卒

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。

#include <stdio.h>
#define N 24typedef 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;
}

#include <stdio.h>
#include <string.h>
#define N 24typedef 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0memset(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;
}