code forces Codeforces Round #487 (Div. 2) C

C. A Mist of Florescence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As the boat drifts down the river, a wood full of blossoms shows up on the riverfront.

"I've been here once," Mino exclaims with delight, "it's breathtakingly amazing."

"What is it like?"

"Look, Kanno, you've got your paintbrush, and I've got my words. Have a try, shall we?"

There are four kinds of flowers in the wood, Amaranths, Begonias, Centaureas and Dianthuses.

The wood can be represented by a rectangular grid of $\mathrm{nn rows\; and mm columns.\; In\; each\; cell\; of\; the\; grid,\; there\; is\; exactly\; one\; type\; of\; flowers.}$

According to Mino, the numbers of connected components formed by each kind of flowers are $\mathrm{aa, bb, cc and dd respectively.\; Two\; cells\; are\; considered\; in\; the\; same\; connected\; component\; if\; and\; only\; if\; a\; path\; exists\; between\; them\; that\; moves\; between\; cells\; sharing\; common\; edges\; and\; passes\; only\; through\; cells\; containing\; the\; same\; flowers.}$

You are to help Kanno depict such a grid of flowers, with $\mathrm{nn and mm arbitrarily\; chosen\; under\; the\; constraints\; given\; below.\; It\; can\; be\; shown\; that\; at\; least\; one\; solution\; exists\; under\; the\; constraints\; of\; this\; problem.}$

Note that you can choose arbitrary $\mathrm{nn and mm under\; the\; constraints\; below,\; they\; are\; not\; given\; in\; the\; input.}$

Input

The first and only line of input contains four space-separated integers $\mathrm{aa, bb, cc and dd (1\le a,b,c,d\le 1001\le a,b,c,d\le 100) —\; the\; required\; number\; of\; connected\; components\; of\; Amaranths,\; Begonias,\; Centaureas\; and\; Dianthuses,\; respectively.}$

Output

In the first line, output two space-separated integers $\mathrm{nn and mm (1\le n,m\le 501\le n,m\le 50) —\; the\; number\; of\; rows\; and\; the\; number\; of\; columns\; in\; the\; grid\; respectively.}$

Then output $\mathrm{nn lines\; each\; consisting\; of mm consecutive\; English\; letters,\; representing\; one\; row\; of\; the\; grid.\; Each\; letter\; should\; be\; among\; \text{'}A\text{'},\; \text{'}B\text{'},\; \text{'}C\text{'}\; and\; \text{'}D\text{'},\; representing\; Amaranths,\; Begonias,\; Centaureas\; and\; Dianthuses,\; respectively.}$

In case there are multiple solutions, print any. You can output each letter in either case (upper or lower).

Examples

input

Copy

5 3 2 1

output

Copy

4 7
DDDDDDD
DABACAD
DBABACD
DDDDDDD

input

Copy

50 50 1 1

output

Copy

4 50
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ABABABABABABABABABABABABABABABABABABABABABABABABAB
BABABABABABABABABABABABABABABABABABABABABABABABABA
DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

input

Copy

1 6 4 5

output

Copy

7 7
DDDDDDD
DDDBDBD
DDCDCDD
DBDADBD
DDCDCDD
DBDBDDD
DDDDDDD

Note

In the first example, each cell of Amaranths, Begonias and Centaureas forms a connected component, while all the Dianthuses form one.

 题意： 告诉你a，b，c，d的个数，要你输出包含这么多个a，b，c，d个数的图

规则：与相同块相联通的是同一个

题解：先构造一个只含有A、B的图，然后i+=2，间隔着往里面加B、C、A、D

QAQ 还是不太会构造题，hzwer了解一下？

QAQorz天下第一

#include<bits/stdc++.h>
using namespace std;
const int M = 48;
const int maxn=64;
char mp[maxn][maxn];

int a,b,c,d;
int main(){
    ios::sync_with_stdio(false);
    while(cin>>a>>b>>c>>d){
        for(int i=0;i<M/2;i++){
            for(int j=0;j<M;j++){
                mp[i][j]='A';
            }
        }
        for(int i=M/2;i<M;i++){
            for(int j=0;j<M;j++){
                mp[i][j]='B';
            }
        }
        a--,b--;
        for(int i=0;i<M/2;i+=2){
            for(int j=0;j<M;j+=2){
                if(b) mp[i][j]='B',b--;
                else if(c) mp[i][j]='C',c--;
            }
        }
        for(int i=M/2+1;i<M;i+=2){
            for(int j=0;j<M;j+=2){
                if(a) mp[i][j]='A',a--;
                else if(d) mp[i][j]='D',d--;
            }
        }
        printf("48 48
");
        for(int i=0;i<M;i++){
            printf("%s
",mp[i]);
        }
    }
    return 0;
}