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Codeforces Round #FF (Div. 2) 题解
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 xi, DZY will put it into the bucket numbered h(xi), 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.
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 xi (0?≤?xi?≤?109).
Output a single integer — the answer to the problem.
10 5 0 21 53 41 53
4
5 5 0 1 2 3 4
-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;
}DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter c DZY knows its value wc. 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?
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 wa to wz. Each such number is non-negative and doesn‘t exceed 1000.
Print a single integer — the largest possible value of the resulting string DZY could get.
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
41
In the test sample DZY can obtain "abcbbc", value?=?1·1?+?2·2?+?3·2?+?4·2?+?5·2?+?6·2?=?41.
链接：http://codeforces.com/contest/447/problem/B
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。
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;
}DZY has a sequence a, consisting of n integers.
We‘ll call a sequence ai,?ai?+?1,?...,?aj (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.
The first line contains integer n (1?≤?n?≤?105). The next line contains n integers a1,?a2,?...,?an (1?≤?ai?≤?109).
In a single line print the answer to the problem — the maximum length of the required subsegment.
6 7 2 3 1 5 6
5
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;
}