Analysis: Cow Race by Steven Hao

From the given constraints, we know that the maximum total time run by either cow is a million (10^6). We can then calculate the position of each cow at every single possible time. Using these positions, we can determine the relative positions of the two cows at any single time. Counting the number of leadership changes is then only a matter of iterating over the time.
Below is a solution in C++ that stores the speeds of both cows at every possible time. 

#include <cstdio>

using namespace std;

const int MAXT = 1001000;
int bspeed[MAXT];
int espeed[MAXT];

int N, M;
int main() {
	freopen("cowrace.in", "r", stdin);
	freopen("cowrace.out", "w", stdout);

	scanf("%d %d", &N, &M);

	int curt = 0;
	for(int i = 0; i < N; ++i) {
		int speed, time;
		scanf("%d %d", &speed, &time);
		for(int j = 0; j < time; ++j) {
			bspeed[curt] = speed;
			++curt;
		}
	}

	curt = 0;
	for(int i = 0; i < M; ++i) {
		int speed, time;
		scanf("%d %d", &speed, &time);
		for(int j = 0; j < time; ++j) {
			espeed[curt] = speed;
			++curt;
		}
	}

	int ans = 0;
	int leader = 0; // 1 if bessie is ahead (or previously ahead, but now  
tied), -1 if elsie ahead (or previously ahead, but now tied).

	int bpos = 0;
	int epos = 0;

	for(int t = 0; t < MAXT; ++t) {
		bpos += bspeed[t];
		epos += espeed[t];
		if (bpos > epos) {
			if (leader == -1) { // leader was elsie
				++ans;
			}
			leader = 1;
		} else if (epos > bpos) {
			if (leader == 1) {
				++ans;
			}
			leader = -1;
		}
	}

	printf("%d\n", ans);
	return 0;
}