A. Find Square
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Consider a table of size n×m
, initially fully white. Rows are numbered 1 through n from top to bottom, columns 1 through m
from left to right. Some square inside the table with odd side length was painted black. Find the center of this square.
Input
The first line contains two integers n
and m (1≤n,m≤115
) — the number of rows and the number of columns in the table.
The i
-th of the next n lines contains a string of m characters si1si2…sim (sij is 'W' for white cells and 'B' for black cells), describing the i
-th row of the table.
Output
Output two integers r
and c (1≤r≤n, 1≤c≤m
) separated by a space — the row and column numbers of the center of the black square.
Examples
Input
Copy
5 6 WWBBBW WWBBBW WWBBBW WWWWWW WWWWWW
Output
Copy
2 4
Input
Copy
3 3 WWW BWW WWW
Output
Copy
2 1
思路：找到第一个B和最后一个B。分别保存下横纵坐标，加起来/2就可以了。
代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,m;
char ch;
int I,J;
int II,JJ;
int lala=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>ch;
if( ch=='B' && lala==0 ){
I=i;
J=j;
lala=100;
}
if(ch=='B'){
II=i;
JJ=j;
}
}
}
/// cout<<II<<" "<<JJ<<endl;
///cout<<I<<" ** "<<J<<endl;
int a=(I+II)/2;
int b=(J+JJ)/2;
cout<<a<<" "<<b<<endl;
return 0;
}B：
这道题还是比较考虑思维的。
http://codeforces.com/contest/1028/problem/B
B. Unnatural Conditions
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Let s(x)
be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b
such that
- s(a)≥n
- ,
- s(b)≥n
- ,
- s(a+b)≤m
- .
Input
The only line of input contain two integers n
and m (1≤n,m≤1129
).
Output
Print two lines, one for decimal representation of a
and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230
.
Examples
Input
Copy
6 5
Output
Copy
6 7
Input
Copy
8 16
Output
Copy
35 53
Note
In the first sample, we have n=6
and m=5. One valid solution is a=6, b=7. Indeed, we have s(a)=6≥n and s(b)=7≥n, and also s(a+b)=s(13)=4≤m.
思路：构造。
代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,m;
/// scanf("%d%d",&n,&m);
for(int i=1;i<=1129;i++){
cout<<1;
}
cout<<endl;
for(int i=1;i<=1128;i++){
cout<<8;
}
cout<<9<<endl;
return 0;
}C:https://blog.csdn.net/xianpingping/article/details/82193032