1890 점프

BOJ::1890 점프

https://www.acmicpc.net/problem/1890

정답에 long형으로 출력해야 한다.

<JAVA>

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
 
public class Main {
    
    static int map[][]=new int[101][101];
    static long d[][]=new long[101][101];
    static int n,cnt=0;
    public static void main(String args[]) throws NumberFormatException, IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        n=Integer.parseInt(br.readLine());
        StringTokenizer st = null;
        
        for(int i = 1 ; i <= n ; i++){
            st = new StringTokenizer(br.readLine());
            for(int j =1 ; j<=n ; j++){
                map[i][j]=Integer.parseInt(st.nextToken());
            }
        }
        
        d[1][1]=1;
        int x;
        for(int i = 1 ; i <= n ; i++){
            for(int j =1 ; j<= n ; j++){
                if(i==n && j==n){
                    continue;
                }
                if(d[i][j]!=0){
                    x = map[i][j];
                    //오른쪽방향
                    if(i+x <=n){
                        d[i+x][j]+=d[i][j];
                    }
                    //아래방향
                    if(j+x<=n){
                        d[i][j+x]+=d[i][j];
                    }
                }
            }
        }
        System.out.println(d[n][n]);
    }
}
 

<C++>

#include <iostream>
using namespace std;
 
int n;
int map[101][101];
long d[101][101];
 
int main() {
    cin >> n;
 
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> map[i][j];
        }
    }
 
    d[1][1] = 1;
    int next;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (d[i][j] == 0 || (i == n && j == n)) {
                continue;
            }
 
            next = map[i][j];
 
            if (next + i <= n) {
                d[next + i][j] += d[i][j];
            }
            if (next + j <= n) {
                d[i][next + j] += d[i][j];
            }
        }
    }
    cout << d[n][n] << endl;
 
}