(Analysis by Nick Wu)

In this problem, we have some notes about how Bessie's, Elsie's, and Mildred's milk outputs change over time. We wish to keep track of the number of days that the cows with the highest milk outputs change.

To compute this information, we need to maintain the milk outputs for each cow for each day when there is a change. At the beginning, we know that each cow outputs exactly 7 gallons of milk. We can go over the notes and figure out if any cow has a differing milk output. After doing so, we can directly determine which cows produced the most milk, and change if there were any changes.

Due to the small number of notes and days, it is not necessary to order the notes by day beforehand.

import java.io.*;
import java.util.*;
public class measurement {
public static void main(String[] args) throws IOException {
// initialize file I/O
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("measurement.out")));

// read in all of the notes
int[] day = new int[n];
String[] cow = new String[n];
int[] change = new int[n];
for(int i = 0; i < n; i++) {
day[i] = Integer.parseInt(st.nextToken());
cow[i] = st.nextToken();
change[i] = Integer.parseInt(st.nextToken());
}

// the milk variables track the amount of milk that each cows was last known to produce
int bessieMilk = 7, elsieMilk = 7, mildredMilk = 7;
// the on variables are true if that cow produced the highest amount of milk on the previous day
boolean bessieOn = true, elsieOn = true, mildredOn = true;

for(int currDay = 1; currDay <= 100; currDay++) {
// look through the notes to see if there were any changes on this day
for(int i = 0; i < n; i++) {
if(day[i] == currDay) {
if(cow[i].equals("Bessie")) {
bessieMilk += change[i];
}
if(cow[i].equals("Elsie")) {
elsieMilk += change[i];
}
if(cow[i].equals("Mildred")) {
mildredMilk += change[i];
}
}
}
// compute the highest milk total and see which cows produced the most milk
int highestMilk = Math.max(bessieMilk, Math.max(elsieMilk, mildredMilk));
boolean bessieOnNext = bessieMilk == highestMilk;
boolean elsieOnNext = elsieMilk == highestMilk;
boolean mildredOnNext = mildredMilk == highestMilk;
if(bessieOn != bessieOnNext || elsieOn != elsieOnNext || mildredOn != mildredOnNext) {
}
bessieOn = bessieOnNext;
elsieOn = elsieOnNext;
mildredOn = mildredOnNext;
}

pw.close();
}

}

For those who prefer C++, here is Brian Dean's solution:
#include <iostream>
#include <fstream>
const int MAX_N = 100;
using namespace std;

// changes[c][d] is the change in milk rate for cow c on day d
// rates[c][d] is the milk rate for cow c on day d
int changes[MAX_N+1];
int rates[MAX_N+1];

// Is cow c the highest on day d?
bool is_highest(int c, int d)
{
int highest = max(max(rates[d], rates[d]), rates[d]);
return rates[c][d] == highest;
}

int main(void)
{
ifstream fin ("measurement.in");
ofstream fout ("measurement.out");

int N, d, c, x;
string name;

fin >> N;
for (int i=0; i<N; i++) {
fin >> d >> name >> x;
if (name == "Bessie") c = 0;
if (name == "Elsie") c = 1;
if (name == "Mildred") c = 2;
changes[c][d] = x;
}

for (int c=0; c<3; c++)
rates[c] = 7;

for (int c=0; c<3; c++)
for (int d=1; d<=100; d++)
rates[c][d] = rates[c][d-1] + changes[c][d];

int num_changes = 0;

for (int d=1; d<=100; d++) {
if (is_highest(0,d-1) != is_highest(0,d) ||
is_highest(1,d-1) != is_highest(1,d) ||
is_highest(2,d-1) != is_highest(2,d))
num_changes++;
}

fout << num_changes << "\n";
return 0;
}