Codeforces Round #316 (Div. 2) A B C

A. Elections

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The country of Byalechinsk is running elections involving
n candidates. The country consists of m cities. We know how many people in each city voted for each candidate.

The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got
the maximum number of votes, then the winner is the one with a smaller index.

At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the
one with a smaller index.

Determine who will win the elections.

Input

The first line of the input contains two integers n,
m (1 ≤ n, m ≤ 100) — the number of candidates and of cities, respectively.

Each of the next m lines contains
n non-negative integers, the j-th number in the
i-th line a_ij (1 ≤ j ≤ n,
1 ≤ i ≤ m,
0 ≤ a_ij ≤ 10⁹) denotes the number of votes for candidate
j in city i.

It is guaranteed that the total number of people in all the cities does not exceed
10⁹.

Output

Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.

Sample test(s)

Input

3 3
1 2 3
2 3 1
1 2 1

Output

2

Input

3 4
10 10 3
5 1 6
2 2 2
1 5 7

Output

1

Note

Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.

Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due
to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.

题意非常坑，大意就是每一个城市给每一个候选人投票，行是城市，列是候选人。每次投票仅仅选最大的，而且编号小的，对于候选人票数同样的情况，也一样，选编号小的。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
typedef long long ll;
int main()
{
    int  n,m;
    ll a[105][105];
    int b[105];
    while(cin>>m>>n)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int i,j;
        for(i=1; i<=n; i++)
            for(j=1; j<=m; j++)
            {
                cin>>a[i][j];
            }
        int k;
        int max1;
        for(i=1; i<=n; i++)
        {
             max1=-1;
            for(j=1; j<=m; j++)
            {
                if(a[i][j]>max1)
                {
                    max1=a[i][j];
                    k=j;
                }
            }
            b[k]++;
        }
        max1=-1;
        for(i=1; i<=100; i++)
            if(b[i]>max1)
                {
                    k=i;
                    max1=b[i];
                }
        cout<<k<<endl;
    }
    return 0;
}

B. Simple Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from
1 to n. Let's assume that Misha chose number
m, and Andrew chose number
a.

Then, by using a random generator they choose a random integer
c in the range between 1 and
n (any integer from 1 to
n is chosen with the same probability), after which the winner is the player, whose number was closer to
c. The boys agreed that if
m and a are located on the same distance from
c, Misha wins.

Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number
n. You need to determine which value of
a Andrew must choose, so that the probability of his victory is the highest possible.

More formally, you need to find such integer a (1 ≤ a ≤ n), that the probability that
is maximal, where
c is the equiprobably chosen integer from
1 to n (inclusive).

Input

The first line contains two integers n and
m (1 ≤ m ≤ n ≤ 10⁹) — the range of numbers in the game, and the number selected by Misha respectively.

Output

Print a single number — such value a, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.

Sample test(s)

Input

3 1

Output

2

Input

4 3

Output

2

Note

In the first sample test: Andrew wins if c is equal to
2 or 3. The probability that Andrew wins is
2 / 3. If Andrew chooses a = 3, the probability of winning will be
1 / 3. If a = 1, the probability of winning is
0.

In the second sample test: Andrew wins if c is equal to
1 and 2. The probability that Andrew wins is
1 / 2. For other choices of
a the probability of winning is less.

yy出来的代码- -

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
typedef long long ll;
int main()
{
    int  n,m;
    while(cin>>n>>m)
    {
        if(n==1 &&m==1)
         {
             cout<<1<<endl;
             continue;
         }
        if(n>=m)
        {
            if(n/2>=m)
            m+=1;
            else
            m-=1;
            cout<<m<<endl;
        }
        else
        {
        if(m/2>=n)
          n+=1;
          else
          n-=1;
          cout<<n<<endl;
        }
    }
    return 0;
}

C. Replacement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of
replacement as the following sequence of steps: find a substring ".." (two consecutive periods) in string
s, of all occurrences of the substring let's choose the first one, and replace this substring with string ".". In other words, during the replacement operation, the first two consecutive
periods are replaced by one. If string s contains no two consecutive periods, then nothing happens.

Let's define f(s) as the minimum number of operations of
replacement to perform, so that the string does not have any two consecutive periods left.

You need to process m queries, the
i-th results in that the character at position
x_i (1 ≤ x_i ≤ n) of string
s is assigned value
c_i. After each operation you have to calculate and output the value of
f(s).

Help Daniel to process all queries.

Input

The first line contains two integers n and
m (1 ≤ n, m ≤ 300 000) the length of the string and the number of queries.

The second line contains string s, consisting of
n lowercase English letters and period signs.

The following m lines contain the descriptions of queries. The
i-th line contains integer
x_i and
c_i (1 ≤ x_i ≤ n,
c_i — a lowercas English letter or a period sign), describing the query of assigning symbol
c_i to position
x_i.

Output

Print m numbers, one per line, the
i-th of these numbers must be equal to the value of
f(s) after performing the i-th assignment.

Sample test(s)

Input

10 3
.b..bz....
1 h
3 c
9 f

Output

4
3
1

Input

4 4
.cc.
2 .
3 .
2 a
1 a

Output

1
3
1
1

Note

Note to the first sample test (replaced periods are enclosed in square brackets).

The original string is ".b..bz....".

• after the first query f(hb..bz....) = 4    ("hb[..]bz...."
   →  "hb.bz[..].."
   →  "hb.bz[..]."
   →  "hb.bz[..]"
   →  "hb.bz.")
• after the second query f(hbс.bz....) = 3    ("hbс.bz[..].."
   →  "hbс.bz[..]."
   →  "hbс.bz[..]"
   →  "hbс.bz.")
• after the third query f(hbс.bz..f.) = 1    ("hbс.bz[..]f."
   →  "hbс.bz.f.")

Note to the second sample test.

The original string is ".cc.".

• after the first query: f(..c.) = 1    ("[..]c."
   →  ".c.")
• after the second query: f(....) = 3    ("[..].."
   →  "[..]."
   →  "[..]"  →  ".")
• after the third query: f(.a..) = 1    (".a[..]"
   →  ".a.")
• after the fourth query: f(aa..) = 1    ("aa[..]"
   →  "aa.")

题意：连续两个点算一个贡献。问的是有多少个贡献。

算出总贡献。我们考虑改变的字符的前后的情况，推断是否加减。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
typedef long long ll;
int main()
{
    int n,m;
    char s[300005];
    char s1[2];
    while(cin>>n>>m)
    {
        scanf("%s",s+1);
        int i;
        int x;
        int l=strlen(s+1);
        int ans=0;
        for(i=1; i<=l; i++)
        {
            if(s[i]=='.'&&s[i+1]=='.')
                ans++;
        }
        while(m--)
        {
            cin>>x>>s1[0];
            if((s1[0]=='.'&&s[x]=='.' )||(s1[0]!='.'&&s[x]!='.'))
            {
                cout<<ans<<endl;
                continue;
            }
            s[x]=s1[0];
            if(s1[0]!='.' &&s[x-1]=='.')
                ans--;
            if(s1[0]!='.' &&s[x+1]=='.')
                ans--;
            if(s1[0]=='.' &&s[x+1]=='.')
                ans++;
            if(s1[0]=='.'&&s[x-1]=='.')
                ans++;
            cout<<ans<<endl;
        }
    }
    return 0;
}