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A. Brain's Photos
time limit per test
memory limit per test
input
output
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
single
n × m, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6
- 'C' (cyan)
- 'M' (magenta)
- 'Y' (yellow)
- 'W' (white)
- 'G' (grey)
- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input
n and m (1 ≤ n, m ≤ 100) — the number of photo pixel matrix rows and columns respectively.
n lines describing matrix rows follow. Each of them contains m
Output
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Examples
input
2 2 C M Y Y
output
#Color
input
3 2 W W W W B B
output
#Black&White
input
1 1 W
output
#Black&White
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m;
char a[10010];
int main()
{
while(~scanf("%d %d",&n,&m))
{
bool flag=0;
for(int i=0;i<n*m;i++)
{
cin>>a[i];
if(a[i]=='C'||a[i]=='M'||a[i]=='Y')
{
flag=1;
}
}
if(flag) puts("#Color");
else puts("#Black&White");
}
return 0;
}
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B. Bakery
time limit per test
memory limit per test
input
output
n cities numbered from 1 to n. There are m
k storages, located in different cities numbered a1, a2, ..., ak.
n - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1
x roubles, if she will open the bakery in some city b (ai ≠ b for every 1 ≤ i ≤ k) and choose a storage in some city s (s = aj for some 1 ≤ j ≤ k) and b and s are connected by some path of roads of summary length x
k
Input
n, m and k (1 ≤ n, m ≤ 105, 0 ≤ k ≤ n) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.
m lines follow. Each of them contains three integers u, v and l (1 ≤ u, v ≤ n, 1 ≤ l ≤ 109, u ≠ v) meaning that there is a road between cities u and v of length of l
k > 0, then the last line of the input contains k distinct integers a1, a2, ..., ak (1 ≤ ai ≤ n) — the number of cities having flour storage located in. If k = 0 then this line is not presented in the input.
Output
Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.
n cities, print - 1
Examples
input
5 4 2 1 2 5 1 2 3 2 3 4 1 4 10 1 5
output
3
input
3 1 1 1 2 3 3
output
-1
Note
Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.
题意：给你 n 个城市，m 条无向边,，k 个有卖面粉的城市，你要在没有卖面粉的地方建一家面包店，每条无向边有一个权值，求从面包店到一个卖面粉的地方花费的最小权值是多少；若不存在到一个卖面粉的城市路，就输出 -1
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAX=1e5+10;
int n,m,k;
bool vis[MAX];
int u[MAX],v[MAX],l[MAX];
int main()
{
while(~scanf("%d %d %d",&n,&m,&k))
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=m;i++)
scanf("%d %d %d",&u[i],&v[i],&l[i]);
int a;
while(k--)
{
scanf("%d",&a);
vis[a]=1; // 标记有卖面粉的城市
}
int ans=INF;
for(int i=1;i<=m;i++)
{
if(vis[u[i]]!=vis[v[i]])
{
ans=min(ans,l[i]);
}
}
if(ans==INF) puts("-1");
else printf("%d\n",ans);
}
return 0;
}
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C. Pythagorean Triples
time limit per test
memory limit per test
input
output
Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are calledPythagorean triples.
For example, triples (3, 4, 5), (5, 12, 13) and (6, 8, 10)
Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.
Katya had no problems with completing this task. Will you do the same?
Input
The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.
Output
Print two integers m and k (1 ≤ m, k ≤ 1018), such that n, m and k
In case if there is no any Pythagorean triple containing integer n, print - 1
Examples
input
3
output
4 5
input
6
output
8 10
input
1
output
-1
input
17
output
144 145
input
67
output
2244 2245
Note
Illustration for the first sample.
题意：给你一个数，让你输出两个数，满足让他们构成勾股数；如果不存在输出 -1
题解：上初中的时候也没听老师说过勾股数有这个性质啊；还是网上看了别人总结的才知道。注意：答案是不唯一的，如输入 9，输出 12，15正确；输出 40，41也正确。
经过分析，可以发现当 a > 1 并且 a 为奇数的时候，把 a 分解成一半 n = ( a - 1 ) / 2；则 b = 2 * n * ( n + 1 )，c = 2 * n * ( n + 1 ) + 1；当 a > 2 并且 a 为偶数的时候,把 a 分解成一半 n = a / 2；则 b = n * n - 1，c = n * n + 1；
#include<cstdio>
#include<algorithm>
#include<cstring>
#define Int __int64
using namespace std;
Int n;
int main()
{
while(~scanf("%I64d",&n))
{
if(n<=2) puts("-1");
else
{
Int m=n>>1;
if(n&1)
printf("%I64d %I64d\n",2*m*(m+1),2*m*(m+1)+1);
else
printf("%I64d %I64d\n",m*m-1,m*m+1);
}
}
return 0;
}
