Codeforces Round #FF (Div. 2) 题解

比赛链接：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 table with p buckets, numbered from 0 to p ?-?1 . He

比赛链接：http://codeforces.com/contest/447

time limit per test

memory limit per test

input

output

DZY has a hash table with p buckets, numbered from 0 to p?-?1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number x_i, DZY will put it into the bucket numbered h(x_i), where h(x) is the hash function. In this problem we will assume, that h(x)?=?x mod p. Operation a mod b denotes taking a remainder after division a by b.

However, each bucket can contain 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 after the i-th insertion, you should output i. If no conflict happens, just output -1.

Input

The first line contains two integers, p and n (2?≤?p,?n?≤?300). Then n lines follow. The i-th of them contains an integer x_i (0?≤?x_i?≤?10⁹).

Output

output a single integer — the answer to the problem.

Sample test(s)

input

10 5
0
21
53
41
53

output

4

input

5 5
0
1
2
3
4

output

-1

链接：http://codeforces.com/contest/447

题意:找出hash时第一次产生冲突的位置。

解题思路：用一个数组表示是否存放有hash之后的元素，0表示这个位置还没有使用过，1表示这个位置上有hash之后的元素（即产生了冲突）。

代码：

#include <cstdio>
#include <cstring>

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

memory limit per test

input

output

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter c DZY knows its value w_c. For each special string s?=?s₁s₂... s_|s| (|s| is the length of the string) he represents its value with a function f(s), where

Now DZY has a string s. He wants to insert k lowercase 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 could get?

Input

The first line contains a single string s (1?≤?|s|?≤?10³).

The second line contains a single integer k (0?≤?k?≤?10³).

The third line contains twenty-six integers from w_a to w_z. 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 could get.

Sample test(s)

input

abc
3
1 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.

题意：在给出的字符串中增加k个字符，求可以得到的最大权值和。

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解题思路：找出最大的位权，将k个有最大位权的字符放到原字符串的末尾，求权值和。ps：答案可能会超出int，要用long long

代码：

#include <iostream>
#include <string>
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

memory limit per test

input

output

DZY has a sequence a, consisting of n integers.

We'll call a sequence a_i,?a_i?+?1,?...,?a_j (1?≤?i?≤?j?≤?n) a subsegment of the sequence a. The value (j?-?i?+?1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, 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 integer n (1?≤?n?≤?10⁵). The next line contains n integers a₁,?a₂,?...,?a_n (1?≤?a_i?≤?10⁹).

Output

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

Sample test(s)

input

6
7 2 3 1 5 6

output

5

Note

You can choose subsegment a₂,?a₃,?a₄,?a₅,?a₆ and change its 3rd element (that is a₄) to 4.

题意：从一串数字中选出一个子串，可以改变子串中一个数字的值得到一个新的子串，求最大的递增新子串的长度。

解题思路：

将原数组分割成递增的子串，记录下每个子串的开始和结束位置，以及长度。接下来要分几种情况讨论：1.相邻的两个子串改变一个数字之后，可以合并形成新的递增子串，2.相邻的3个子串，中间子串长度为1，改变中间的数字后可以形成新的递增子串，3.相邻的子串不能合并形成新的递增子串，但是可以在原串的基础上，得到一个长度增加1的新的递增子串（在子串开头位置前有数字，或是结束位置后有数字）。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
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;
}