标签:水题 .. characters Enlarge 2019 Taotao each test include
Problem Description Taotao is glad to see that lots of students have registered for today's algorithm competition. Thus he wants to pull a banner on the wall. But Taotao is so stupid that he doesn't even know how to enlarge words to fit the size of banner. So Taotao gives you a slogan template and asked you to enlarge them in k times.Input There are multiple test cases.
For each test case, first line contains three numbers n, m, k (1<=n, m<=100,1<=k<=10).
The next n lines each contain m visible characters.
Output For each test case, you should output k*n lines each contain k*m visible characters to represent the expanded slogan.
Sample Input 3 3 2 .*. .*. .*.
Sample Output ..**.. ..**.. ..**.. ..**.. ..**.. ..**..
#include <cstdio> #include <iostream> #include <vector> #include <string> #include <cstring> #include <iostream> #include <algorithm> #include <queue> using namespace std; #define ll long long const int inf = 0x3f3f3f3f; const ll linf =1LL<<50; const int maxn = 1e5+8; const ll lmaxn = 2*1e9+8; const ll mod = 1000000007; int sign[10000+8]; char a[100+8][100+8]; int n, m, k; int main() { while(~scanf("%d%d%d", &n, &m, &k) && (n+m+k)) { getchar(); int cnt = 0; for(int i = 1; i<=n; i++) { for(int j = 1; j<=m; j++) { scanf("%c", &a[i][j]); if(j>1 && a[i][j] != a[i][j-1]) { sign[cnt++] = j-1; // cout<<sign[cnt-1]<<endl; } if(j == m) { sign[cnt++] = j; // cout<<sign[cnt-1]<<endl; } } getchar(); } int miao = 0, ying = 0;; for(int i = 1; i <= n; i++) { for(int j = 1; j <= m*k; j++) { printf("%c",a[i][sign[miao]]); if(j == sign[miao]*2) { miao++; } } printf("\n"); for(int j = 1; j <= m*k; j++) { printf("%c",a[i][sign[ying]]); if(j == sign[ying]*2) { ying++; } } printf("\n"); } } return 0; }
标签:水题,..,characters,Enlarge,2019,Taotao,each,test,include 来源: https://www.cnblogs.com/RootVount/p/10741494.html
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