2018國慶第七場個人賽
codeforces 719A
Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for n consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains n integers ai (0 ≤ ai ≤ 15) — Vitya's records.
It's guaranteed that the input data is consistent.
Output
If Vitya can be sure that the size of visible part of the moon on day n + 1 will be less than the size of the visible part on day n, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.
Sample Input
Input
5
3 4 5 6 7Output
UP
Input
7
12 13 14 15 14 13 12Output
DOWN
Input
1
8Output
-1
Sample Output
Hint
In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.
看是遞增的還是遞減的，注意幾種情況
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <deque>
#include <set>
using namespace std;
typedef long long ll;
#define MAX 0x3f3f3f
map<ll,ll> M;
set<ll>s;
ll a[1110];
int main()
{
ll n,i,sum;
cin>>n;
for(i=0;i<n;i++)
cin>>a[i];
if(n==1)
{
if(a[0]==15)
cout<<"DOWN"<<endl;
else
if(a[0]==0)
cout<<"UP"<<endl;
else
cout<<"-1"<<endl;
}
else
{
if(a[n-2]==1&&a[n-1]==0)
cout<<"UP"<<endl;
else
if(a[n-1]==15&&a[n-2]==14)
cout<<"DOWN"<<endl;
else
{
if(a[n-2]<a[n-1])
cout<<"UP"<<endl;
else
if(a[n-2]>a[n-1])
cout<<"DOWN"<<endl;
else
cout<<"-1"<<endl;
}
}
return 0;
}
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are n cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of cockroaches.
The second line contains a string of length n, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
Output
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
Sample Input
Input
5 rbbrr
Output
1
Input
5 bbbbb
Output
2
Input
3 rbr
Output
0
給你一個由'r','b'組成的字串，你可以進行兩種操作：（1）把任意兩個位置的字母互換；（2）把任意一個位置的r變成b，或b變成r。
輸入一個字串，問至少變幾次才能成為r,b交錯的序列？（rbrbrb….或者brbrbr…都行）
題目分析：
這是一個貪心策略，不過我一開始沒想到。
分別考慮兩種最終結果，計算字串與目標串中r錯位和b錯位的個數（即目標為b，但實際為r和目標為r和實際為b），設為x1,x2，則要想把該字串變成目標字串，需要操作的次數為max(x1,x2)
.
（想想為什麼？其實道理很簡單，首先通過min(x1,x2)次互換操作，再把剩下的r變成b或b變成r，需要|x1−x2|次操作，這倆加起來就是max(x1,x2).）
最後取兩種目標串操作的最小值就可以了。
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <deque>
#include <set>
using namespace std;
typedef long long ll;
#define MAX 0x3f3f3f
map<ll,ll> M;
set<ll> S;
char s[100100];
int main()
{
ll i,a=0,b=0,n,ans1=0,ans2=0;
cin>>n>>s;
for(i=0;i<n;i++)//brbr……
{
if(i%2==0)
{
if(s[i]=='r')
a++;
}
else
{
if(s[i]=='b')
b++;
}
}
ans1=max(a,b);
a=0,b=0;
for(i=0;i<n;i++)//rbrb……
{
if(i%2==0)
{
if(s[i]=='b')
a++;
}
else
{
if(s[i]=='r')
b++;
}
}
ans2=max(a,b);
cout<<min(ans1,ans2)<<endl;;
return 0;
}
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
Input
The first and only line of the input contains a single string s (1 ≤ |s| ≤ 50 000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
Output
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print - 1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
Sample Input
Input
ABC??FGHIJK???OPQR?TUVWXY?
Output
ABCDEFGHIJKLMNOPQRZTUVWXYS
Input
WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO
Output
-1
Input
??????????????????????????
Output
MNBVCXZLKJHGFDSAQPWOEIRUYT
Input
AABCDEFGHIJKLMNOPQRSTUVW??M
Output
-1
Hint
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer.
給出一個字串，判斷其是否存在一個子串（滿足：包含26個英文字母且不重複，字串中有‘？’表示佔位符可表示字母），如果存在則輸出該字串‘？’位置用替換後的字母代替，其他不在子串中的‘？’用字母代替即可。如果該字串不存在滿足條件的子串，則輸出-1.
還是自己寫一下比較好
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <deque>
#include <set>
using namespace std;
typedef long long ll;
#define MAX 0x3f3f3f
map<ll,ll> M;
set<ll> S;
int main()
{
ll n,i,sum=0,j,k,ans=0;
string s,ss;
cin>>s;
n=s.size();
if(n<26)
{
cout<<"-1"<<endl;
return 0;
}
else
{
for(i=0;i<=n-26;i++)
{
ss=s.substr(i,26);
//cout<<ss<<endl;
M.clear();
S.clear();
sum=0;
for(j=0;j<26;j++)
{
if(ss[j]=='?')
sum++;
else
{
M[ss[j]-'A']++;
S.insert(ss[j]);
}
}
//cout<<sum<<" "<<S.size()<<endl;
if(sum+S.size()==26)
{
for(j=0;j<i;j++)
{
if(s[j]=='?')
cout<<"A";
else
cout<<s[j];
}
for(j=0;j<26;j++)
{
if(ss[j]=='?')
{
for(k=0;k<26;k++)
{
if(M[k]==0)
{
printf("%c",k+'A');
M[k]=1;
break;
}
}
}
else
cout<<ss[j];
}
for(j=i+26;j<n;j++)
{
if(s[j]=='?')
cout<<"A";
else
cout<<s[j];
}
break;
}
else
ans++;
}
if(ans==n-26+1)
cout<<"-1"<<endl;
}
return 0;
}