过河卒 NOIp 2002 dp-CFANZ编程社区

题目描述

棋盘上过河卒 NOIp 2002 dp_#include

棋盘用坐标表示，过河卒 NOIp 2002 dp_#include_02

现在要求你计算出卒从过河卒 NOIp 2002 dp_#include_03

输入输出格式

输入格式：

一行四个数据，分别表示过河卒 NOIp 2002 dp_#define_04

输出格式：

一个数据，表示所有的路径条数。

输入输出样例

输入样例#1：

复制

6 6 3 3

输出样例#1： 复制

说明

结果可能很大！

一个人问我的，那我就干脆写上吧。。

直接无脑dfs40分，

考虑dp即可；

很容易想到 dp[ i ][ j ]=dp[ i-1 ][ j ]+dp[ i ][ j-1 ]；

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#pragma GCC optimize(2)
//#include
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 700005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)

inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
    ll ans = 1;
    a = a % c;
    while (b) {
        if (b % 2)ans = ans * a%c;
        b /= 2; a = a * a%c;
    }
    return ans;
}

int n, m;
int Map[100][100];
ll ans;
int dx[] = { 1,1,2,2,-1,-1,-2,-2 };
int dy[] = { 2,-2,1,-1,2,-2,1,-1 };

int ddx[] = { 1,0 };
int ddy[] = { 0,1 };

bool check(int x, int y) {
    return (x >= 0 && x <= n&&y >= 0 && y <= m);
}

ll dp[100][100];

int main(){
    //ios::sync_with_stdio(0);
    rdint(n); rdint(m);
    int x, y; rdint(x); rdint(y); Map[x][y] = 1;
    for (int i = 0; i < 8; i++) {
        int nx = x + dx[i], ny = y + dy[i];
        if (nx >= 0 && nx <= n&&ny >= 0 && ny <= m)Map[nx][ny] = 1;
    }
    dp[0][0] = 0; dp[1][0] = 1; dp[0][1] = 1;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            if (Map[i][j])continue;
            if (!Map[i][j + 1]) {
                dp[i][j + 1] += dp[i][j];
            }
            if (!Map[i + 1][j]) {
                dp[i + 1][j] += dp[i][j];
            }
        }
    }
    cout << dp[n][m] << endl;
    return 0;
}

EPFL - Fighting