CodeforcesRound # FF (Div.2) question

Link: codeforces. comcontest447A. Large, numberedfrom0top? -? 1. He

Match link: http://codeforces.com/contest/447. DZY Loves Hash time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY has a hash table with p buckets, numbered from 0 to p? -? 1. He

Match link: http://codeforces.com/contest/447

A. DZY Loves Hash

Time limit per test

1 second

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

DZY has a hash tablePBuckets, numbered from 0P? -? 1. He wants to insertNNumbers, in the order they are given, into the hash table. ForI-Th numberXI, DZY will put it into the bucket numberedH(XI), WhereH(X) Is the hash function. In this problem we will assume, thatH(X)? =?X mod p. OperationA mod BDenotes taking a remainder after pisionAByB.

However, each bucket can contain in no more than one element. if DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. suppose the first conflict happens right afterI-Th insertion, you shoshould outputI. If no conflict happens, just output-1.

Input

The first line contains two integers,PAndN(2? ≤?P,?N? ≤? 300). ThenNLines follow.I-Th of them contains an integerXI(0? ≤?XI? ≤? 109 ).

Output

Output a single integer-the answer to the problem.

Sample test (s)

Input

10 5021534153

Output

4

Input

5 501234

Output

-1

Link: http://codeforces.com/contest/447

Find the location where hash conflicts occur for the first time.

Solution: use an array to indicate whether the elements after hash are stored. 0 indicates that this position has not been used, 1 indicates that there is a hash element in this position (that is, a conflict occurs ).

Code:

#include 
 
  #include 
  
   const int MAXN = 305;int a[MAXN], p, n, ans = -1;int main(){    bool flag = true;    scanf("%d%d", &p, &n);    for(int i = 0; i < n; i++)    {        int x;        scanf("%d", &x);        if(flag)        {            if(0 == a[x % p])            {                a[x % p] = 1;            }            else            {                ans = i + 1;                flag = false;            }        }    }    printf("%d\n", ans);    return 0;}
  
 

B. DZY Loves Strings

Time limit per test

1 second

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letterCDZY knows its valueWC. For each special stringS? =?S1S2...S|S| (|S| Is the length of the string) he represents its value with a functionF(S), Where

Now DZY has a stringS. He wants to insertKLowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he cocould get?

Input

The first line contains a single stringS(1? ≤? |S|? ≤? 103 ).

The second line contains a single integerK(0? ≤?K? ≤? 103 ).

The third line contains twenty-six integers fromWAToWZ. Each such number is non-negative and doesn' t exceed 1000.

Output

Print a single integer-the largest possible value of the resulting string DZY cocould get.

Sample test (s)

Input

abc31 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Output

41

Note

In the test sample DZY can obtain "abcbbc ",Value? =? 1-1? +? 2 · 2? +? 3 · 2? +? 4 · 2? +? 5 · 2? +? 6 · 2? =? 41.

Link: http://codeforces.com/contest/447/problem/ B

Add k characters to the given string to obtain the maximum weight and sum.

Solution: Find the maximum bit right, put k characters with the maximum bit right at the end of the original string, and evaluate the sum of values. Ps: The answer may exceed int. Use long.

Code:

#include 
 
  #include 
  
   using namespace std;int main(){    string s;    int k, a[27], imax = -1;    cin >> s >> k;    for(int i = 0; i < 26; i++)    {        cin >> a[i];        if(a[i] > imax)        {            imax = a[i];        }    }    long long ans = 0;    int len = s.length();    for(int i = 0; i < len; i++)    {        ans += a[s[i] - 'a'] * (i + 1);    }    ans += (long long)imax * k * (2 * len + k + 1) / 2;    cout << ans << endl;}
  
 

C. DZY Loves Sequences

Time limit per test

1 second

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

DZY has a sequenceA, ConsistingNIntegers.

We'll call a sequenceAI,?AI? +? 1 ,?...,?AJ(1? ≤?I? ≤?J? ≤?N) A subsegment of the sequenceA. The value (J? -?I? +? 1) denotes the length of the subsegment.

Your task is to find the longest subsegmentA, Such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integerN(1? ≤?N? ≤? (105). The next line containsNIntegersA1 ,?A2 ,?...,?AN(1? ≤?AI? ≤? 109 ).

Output

In a single line print the answer to the problem-the maximum length of the required subsegment.

Sample test (s)

Input

67 2 3 1 5 6

Output

5

Note

You can choose subsegmentA2 ,?A3 ,?A4 ,?A5 ,?A6 and change its 3rd element (that isA4) to 4.

Link: http://codeforces.com/contest/447/problem/C

Select a substring from a number to change the value of a number in the substring to a new substring, and increase the length of the new substring to the maximum.

Solution:

Splits the original array into ascending substrings and records the start and end positions and lengths of each substring. Next we will discuss the situation in several ways: 1. after two adjacent substrings change a number, they can be merged to form a new incrementing substring. the length of the adjacent three substrings is 1. After the numbers in the middle are changed, a new ascending substring can be formed. adjacent substrings cannot be merged to form new incremental substrings, but can be based on the original strings, get a new incrementing substring with a length increase of 1 (there is a number before the starting position of the substring, or there is a number after the ending position ).

Code:

#include 
 
  #include 
  
   #include using namespace std;const int MAXN = 100010;int a[MAXN];struct P{    int l, len, r;};P p[MAXN];int n;int main(){    memset(p, 0, sizeof(p));    scanf("%d", &n);    int t = 0;    for(int i = 0; i < n; i++)    {        scanf("%d", &a[i]);        if(!i)        {            p[t].len++;            p[t].l = p[t].r = i;            continue;        }        if(a[i] <= a[i - 1])        {            t++;        }        if(0 == p[t].len)        {            p[t].l = i;        }        p[t].len++;        p[t].r = i;    }    int ans = p[0].len < n ? p[0].len + 1 : p[0].len;    for(int i = 1; i <= t; i++)    {        ans = max(ans, p[i].len + 1);        if(a[p[i].l] > a[p[i - 1].r - 1] + 1 ||           a[p[i].l + 1] > a[p[i - 1].r] + 1)        {            ans = max(ans, p[i].len + p[i - 1].len);        }        if(i >= 2 && 1 == p[i - 1].len &&           a[p[i].l] > a[p[i - 2].r + 1])        {            ans = max(ans, p[i].len + p[i - 2].len + 1);        }     //   printf("%d \n", p[i].len);    }    printf("%d\n", ans);    return 0;}