[BOJ/백준][Silver3] 16967 : 배열 복원하기 (C++)

https://www.acmicpc.net/problem/16967

16967번: 배열 복원하기

 

 

[난이도] Silver3
[유형] 구현

[풀이]
두 배열의 겹치는 부분은 Bi,j = Ai,j + Ai-X,j-Y 와 같으므로
겹치는 부분의 A배열의 값은 A[i][j] = B[i][j] - A[i-X][j-Y] 의 식으로 구할 수 있습니다.
A[0][0] 부터 구하기 시작하면 A[i-X][j-Y]는 이미 구해져 있다는 것이 보장되기 때문에
위의 식에서 A[i-X][j-Y]이 없어서 못구하는 상황은 발생하지 않습니다.

 

#include <cstdio>
int H,W,X,Y,B[700][700],A[300][300];
int main(){
    scanf("%d%d%d%d",&H,&W,&X,&Y);
    for(int i=0;i<H+X;i++)
        for(int j=0;j<W+Y;j++) scanf("%d",&B[i][j]);
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            if(i>=X && j>=Y) A[i][j] = B[i][j] - A[i-X][j-Y];
            else A[i][j] = B[i][j];
            printf("%d ",A[i][j]);
        }
        puts("");
    }
}

https://github.com/has2/Problem-Solving/blob/master/boj-solved.ac/Silver3/16967.cpp