X and paintings
X is well known artist, no one knows the secrete behind the beautiful paintings of X except his friend Y, well the reason that Y knows X's secrete is that he is that secret. Y is a programmer and he helps X with drawing paintings using computer program written by him. Unfortunately Y program is not working any more, now it's your turn to help X by writing a program that helps him in painting, your program should accept a sequence of instructions, each will draw an opaque (filled) rectangle in the form: r1, c1, r2, c2, color. where (r1, c1) is the upper left corner of the rectangle and (r2, c2) is the lower right corner, and color is a character that denotes the color of this rectangle. all rectangles will be printed on a R by Cplane, by default this plane is filled with dots (i.e. '.'). R and C between 1 and 100. for each instruction (1 ≤ ri ≤ R) and (1 ≤ ci ≤ C) and color is a ASCII character [a-z]. The number of instructions won't exceed 100 instructions.
Input
The first line of input contains an integer T denotes number of test cases. The first line of each test case contains three integers R, C ,I where R and C denotes number of rows and columns of the painting, and I denotes number of instructions. Each of the next I lines contains four integers r1,c1,r2,c2 and the color character.
Output
Print the final plane after evaluating the instructions in order.
Example
Input
1 5 5 3 1 1 2 2 a 1 2 5 5 c 2 2 3 3 dOutput
acccc addcc .ddcc .cccc .cccc
好喜欢这个点阵输出的提亚马特！
就喜欢这么方的 阵！
上 代码 （代码来自社会我浩哥）jie lian 这里有链接
tip：
1、今天 队友做题， 把s【】【】开成100 100 了， 就rte了，
2、 memset(s, '\0', sizeof(s)); 以后也习惯去用 memset 字符串初四花为 ‘\0’;
3、先是点阵， 再依次更新图阵， 虽然实为水题， 但是很喜欢 游戏四哈。
4、这个题 哈哈哈 内心wa 了无数遍了
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
ios::sync_with_stdio(0);
int t;
cin >> t;
char s[105][105], s1;
while(t--)
{
int r, c, i;
cin >> r >> c >> i;
memset(s, '\0', sizeof(s));
for(int j = 1; j <= r; j++)
{
for(int k = 1; k <= c; k++)
{
s[j][k] = '.';
}
}
int r1, c1, r2, c2;
while(i--)
{
cin >> r1 >> c1 >> r2 >> c2 >> s1;
for(int j = r1; j <= r2; j++)
{
for(int k = c1; k <= c2; k++)
{
s[j][k] = s1;
}
}
}
for(int j = 1; j <= r; j++)
{
for(int k = 1; k <= c; k++)
{
cout << s[j][k];
}
cout << endl;
}
}
return 0;
}