#55540: c++ 解答
方法一: for迴圈直接讓陣列長大即可
#include <bits/stdc++.h>
using namespace std;
int r, c, tr, tc;
int org[15][15];
int main() {
cin >> r >> c >> tr >> tc;
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) cin >> org[i][j];}
for (int i = 0; i < r; i++) {
for (int step_r = 0; step_r < tr; step_r++) {
for (int j = 0; j < c; j++) {
for (int step_c = 0; step_c < tc; step_c++) {
cout << org[i][j] << " ";
}
}
cout << "\n";
}
}
return 0;
}
方法二: 一開始想太複雜寫的遞迴法(可ac 但是很慢)
#include <bits/stdc++.h>
using namespace std;
int r,c,tr,tc;
int nw[105][105];
int org[15][15];
void dfs(int curr,int curc,int limr,int limc,int num){
if(curr>limr||curc>limc||curr<0||curc<0)return;
nw[curr][curc]=num;
dfs(curr+1,curc,limr,limc,num);
nw[curr][curc]=num;
dfs(curr,curc+1,limr,limc,num);
return;
}
int main(){
cin>>r>>c>>tr>>tc;
for(int i=0;i<r;i++){
for(int j=0;j<c;j++)cin>>org[i][j];
}
memset(nw,-1,sizeof(nw));
for(int i=0;i<r*tr;i++){
for(int j=0;j<c*tc;j++){
if(nw[i][j]==-1)dfs(i,j,i+tr-1,j+tc-1,org[i/tr][j/tc]);
}
}
for(int i=0;i<r*tr;i++){
for(int j=0;j<c*tc;j++){cout<<nw[i][j]<<" ";}cout<<endl;}
return 0;
}