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.
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 aij (1 ≤ j ≤ n, 1 ≤ i ≤ m, 0 ≤ aij ≤ 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.
Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.
3 3 1 2 3 2 3 1 1 2 1
2
3 4 10 10 3 5 1 6 2 2 2 1 5 7
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.
#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;
}
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).
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.
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.
3 1
2
4 3
2
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.
#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;
}
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 xi (1 ≤ xi ≤ n) of string s is assigned value ci. After each operation you have to calculate and output the value of f(s).
Help Daniel to process all queries.
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 xi and ci (1 ≤ xi ≤ n, ci — a lowercas English letter or a period sign), describing the query of assigning symbol ci to position xi.
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.
10 3 .b..bz.... 1 h 3 c 9 f
4 3 1
4 4 .cc. 2 . 3 . 2 a 1 a
1 3 1 1
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.")
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;
}