Atcoder CODE FESTIVAL 2017 qual C 总结+ABCD题解

这场比赛..打得还不错..最后两题我想不出..不过还好过了的人也不怎么多。所以我以65分钟过了4题的成绩排在了rank167.(顺带Orz orbitingflea)最终加了42，rating：2026（终于黄名了qwq）..朝着下一个目标进发！

以下是题解（ABCD）

=================我是萌萌哒分割线OvO=================

A - Can you get AC?

Problem Statement

Snuke built an online judge to hold a programming contest.

When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)

Determine whether the judge can return the string “AC” as the verdict to a program.

Constraints

2≤|S|≤5

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If the judge can return the string “AC” as a verdict to a program, print “Yes”; if it cannot, print “No”.

Examples

Example 1

Input

BACD

Output

Yes

The string “AC” appears in “BACD” as a contiguous substring (the second and third characters).

Example 2

Input

ABCD

Output

No

Although the string “ABCD” contains both “A” and “C” (the first and third characters), the string “AC” does not appear in “ABCD” as a contiguous substring.

Example 3

Input

CABD

Output

No

Example 4

Input

ACACA

Output

Yes

Example 5

Input

XX

Output

No

题意

给你一个字符串，问你串中是否包含子串“AC”。

思路

大水题，扫过去判一下就好了..

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>
inline void read(T &x) {
Finish_read=0;x=0;int f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
x*=f;Finish_read=1;
}
template<class T>
inline void print(T x) {
if(x/10!=0)
print(x/10);
putchar(x%10+'0');
}
template<class T>
inline void writeln(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
putchar('\n');
}
template<class T>
inline void write(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
}
/*================Header Template==============*/
string s;
int main() {
cin>>s;
for(int i=0;i<s.length()-1;i++)
if(s[i]=='A'&&s[i+1]=='C') {
puts("Yes");
return 0;
}
puts("No");
return 0;
}

B - Similar Arrays

Problem Statement

We will say that two integer sequences of length N, x1,x2,…,xN and y1,y2,…,yN, are similar when |xi−yi|≤1 holds for all i (1≤i≤N).

In particular, any integer sequence is similar to itself.

You are given an integer N and an integer sequence of length N, A1,A2,…,AN.

How many integer sequences b1,b2,…,bN are there such that b1,b2,…,bN is similar to A and the product of all elements, b1b2…bN, is even?

Constraints

1≤N≤10

1≤Ai≤100

Input

Input is given from Standard Input in the following format:

N

A1 A2 … AN

Output

Print the number of integer sequences that satisfy the condition.

Examples

Example 1

Input

2

2 3

Output

7

There are seven integer sequences that satisfy the condition:

1,2

1,4

2,2

2,3

2,4

3,2

3,4

Example 2

Input

3

3 3 3

Output

26

Example 3

Input

1

100

Output

1

Example 4

Input

10

90 52 56 71 44 8 13 30 57 84

Output

58921

题意

给你一个长度为n的数组，然后让你求与他相似的数组的个数，定义相似为两个数组长度相等且每一位的差的绝对值不超过1，并且数组所有数的乘积不是奇数

思路

其实这题有O(n)做法，只是我看范围也不大，我就写了个DFS..没想到竟然没有T还只用了2ms..233333.对于DFS的话就是每一位扫3个数推下去就行了

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>
inline void read(T &x) {
Finish_read=0;x=0;int f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
x*=f;Finish_read=1;
}
template<class T>
inline void print(T x) {
if(x/10!=0)
print(x/10);
putchar(x%10+'0');
}
template<class T>
inline void writeln(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
putchar('\n');
}
template<class T>
inline void write(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
}
/*================Header Template==============*/
int arr[15],a[15],n,ans=0;
ll sum=1;
inline void dfs(int pos) {
if(pos==n+1) {
sum=1;
for(int i=1;i<=n;i++)
sum*=a[i];
if(sum%2==0)
ans++;
return;
}
for(int i=arr[pos]-1;i<=arr[pos]+1;i++) {
a[pos]=i;
dfs(pos+1);
}
}
int main() {
read(n);
for(int i=1;i<=n;i++)
read(arr[i]);
dfs(1);
writeln(ans);
}

C - Inserting ‘x’

Problem Statement

We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly:

Insert a letter x to any position in s of his choice, including the beginning and end of s.

Snuke’s objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.

Notes

A palindrome is a string that reads the same forward and backward. For example, “a”, “aa”, “abba” and “abcba” are palindromes, while “ab”, “abab” and “abcda” are not.

Constraints

1≤|s|≤105

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

If the objective is achievable, print the number of operations required. If it is not, print “-1” instead.

Examples

Example 1

Input

xabxa

Output

2

One solution is as follows (newly inserted ‘x’ are shown in bold):

xabxa → xa x bxa → xaxbxa x

Example 2

Input

ab

Output

-1

No sequence of operations can turn s into a palindrome.

Example 3

Input

a

Output

0

s is a palindrome already at the beginning.

Example 4

Input

oxxx

Output

3

One solution is as follows:

oxxx → x oxxx → x xoxxx → x xxoxxx

题意

给你一个字符串，问你能不能通过加最少的x来获得回文串。

思路

首先，对于原串，我们将所有的x拿掉，进行一遍回文串判定，如果不是回文串，那么很显然直接输出-1，否则将原串用两个指针i,j分别从右往左和从左往右向中间扫。如果s[i]==s[j]那么直接i++,j–；否则如果s[i]==’x’，那么就ans++,i++；否则就是s[j]==’x’，那么就ans++,j–；最后扫到i>=j就行了。

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>
inline void read(T &x) {
Finish_read=0;x=0;int f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
x*=f;Finish_read=1;
}
template<class T>
inline void print(T x) {
if(x/10!=0)
print(x/10);
putchar(x%10+'0');
}
template<class T>
inline void writeln(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
putchar('\n');
}
template<class T>
inline void write(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
}
/*================Header Template==============*/
string s,n="";
int ans=0;
int main() {
cin>>s;
for(int i=0;i<s.length();i++)
if(s[i]!='x')
n+=s[i];
for(int i=0,j=n.length()-1;i<=j;i++,j--)
if(n[i]!=n[j]) {
puts("-1");
return 0;
}
for(int i=0,j=s.length()-1;i<=j;) {
if(s[i]==s[j]) {
i++;
j--;
continue;
}
if(s[i]=='x') {
ans++;
i++;
}
if(s[j]=='x') {
ans++;
j--;
}
}
printf("%d\n",ans);
}

D - Yet Another Palindrome Partitioning

Problem Statement

We have a string s consisting of lowercase English letters. Snuke is partitioning s into some number of non-empty substrings. Let the subtrings obtained be s1, s2, …, sN from left to right. (Here, s=s1+s2+…+sN holds.) Snuke wants to satisfy the following condition:

For each i (1≤i≤N), it is possible to permute the characters in si and obtain a palindrome.

Find the minimum possible value of N when the partition satisfies the condition.

Constraints

1≤|s|≤2×105

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum possible value of N when the partition satisfies the condition.

Examples

Example 1

Input

aabxyyzz

Output

2

The solution is to partition s as “aabxyyzz” = “aab” + “xyyzz”. Here, “aab” can be permuted to form a palindrome “aba”, and “xyyzz” can be permuted to form a palindrome “zyxyz”.

Example 2

Input

byebye

Output

1

“byebye” can be permuted to form a palindrome “byeeyb”.

Example 3

Input

abcdefghijklmnopqrstuvwxyz

Output

26

Example 4

Input

abcabcxabcx

Output

3

The solution is to partition s as “abcabcxabcx” = “a” + “b” + “cabcxabcx”.

题意

给你一个字符串，让你将他分成最少的子串，使得每个子串重排列后都是一个回文串

思路

考虑dp，f[i]表示到i这一位最少可以分成几个串。对于每一个位置，我们记录从字符串开头到该位置的所有字母的个数是单数还是双数，这样我们可以对于每一个位置就是一个26位的01串，1表示这种字符是奇数，否则是偶数。然后用一个数组记录每种状态f值最小的位置，然后对于转移，我们枚举其中一个字母的数量是奇是偶与当前状态不一样的一个（还有自身原来的），然后求一个f值的min，最后对找到的那个最小值+1就是这个位置的值了。

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>
inline void read(T &x) {
Finish_read=0;x=0;int f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
x*=f;Finish_read=1;
}
template<class T>
inline void print(T x) {
if(x/10!=0)
print(x/10);
putchar(x%10+'0');
}
template<class T>
inline void writeln(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
putchar('\n');
}
template<class T>
inline void write(T x) {
if(x<0)
putchar('-');
x=abs(x);
print(x);
}
/*================Header Template==============*/
const int mx=(1<<26);
int pos[mx],f[200010];
char s[200010];
bitset<26> x[200010];
int main() {
scanf("%s",s+1);
int n=strlen(s+1);
f[0]=0;
memset(pos,-1,sizeof pos);
pos[0]=0;
for(int i=1;i<=n;i++) {
x[i]^=x[i-1];
x[i][s[i]-'a']=1-x[i][s[i]-'a'];
f[i]=2e9;
int t=x[i].to_ulong(),ind=i;
for(int j=0;j<26;j++) {
int tmp=(1<<j)^t;
if(pos[tmp]!=-1&&f[ind]>f[pos[tmp]])
ind=pos[tmp];
}
if(pos[t]!=-1&&f[ind]>f[pos[t]])
ind=pos[t];
f[i]=f[ind]+1;
if(pos[t]==-1||f[i]<f[pos[t]])
pos[t]=i;
}
printf("%d\n",f
);
return 0;
}