NC52863. Super Resolution
描述
输入描述
The input contains zero or more test cases and is terminated by end-of-file. For each test case,
The first line contains four integers n, m, a, b.
The i-th of the following n lines contains a binary string of length m which denotes the i-th row of the original picture. Character "`0`" stands for a white pixel while the character "`1`" stands for black one.
*
* The number of tests cases does not exceed 10.
输出描述
For each case, output rows and columns which denote the result.
示例1
输入:
2 2 1 1 10 11 2 2 2 2 10 11 2 2 2 3 10 11
输出:
10 11 1100 1100 1111 1111 111000 111000 111111 111111
Pascal(fpc 3.0.2) 解法, 执行用时: 2ms, 内存消耗: 220K, 提交时间: 2019-10-04 12:14:21
var a:string; i,j,i1,j1,n,m,n1,m1:longint; begin readln(n,m,n1,m1); while not eof do begin for i:=1 to n do begin readln(a); for i1:=1 to n1 do begin for j:=1 to m do for j1:=1 to m1 do write(a[j]); writeln; end; end; readln(n,m,n1,m1); end; end.
C++14(g++5.4) 解法, 执行用时: 3ms, 内存消耗: 476K, 提交时间: 2019-10-04 14:44:21
#include<bits/stdc++.h> using namespace std; typedef long long ll; char s[15][15]; int main(){ int a,b,c,d; while(~scanf("%d%d%d%d",&a,&b,&c,&d)){ for(int i=0;i<a;i++)scanf("%s",s[i]); for(int i=0;i<a*c;i++){ for(int o=0;o<b*d;o++) printf("%c",s[i/c][o/d]); printf("\n"); } } }
C++11(clang++ 3.9) 解法, 执行用时: 4ms, 内存消耗: 460K, 提交时间: 2020-02-25 21:11:38
#include<bits/stdc++.h> using namespace std; int n,m,a,b,i,j; string s,t; int main() { while(cin>>n>>m>>a>>b) { for(i=0;i<n;i++) { cin>>s; t=""; for(auto c:s) t+=string(b,c); for(j=0;j<a;j++) cout<<t<<endl; } } return 0; }
Python3(3.5.2) 解法, 执行用时: 24ms, 内存消耗: 3420K, 提交时间: 2019-10-04 12:43:01
while 1: try:n,m,a,b=map(int,input().split()) except:break for i in range(n): s=input();S="" for j in s:S+=j*b print(end=(S+'\n')*a)