一般化要考虑数值全部相同到块到情形.
比如
AAAAAA
XXXXXX
BBBBBB
X若超过一行,还需要A=B
同附上AC代码.
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

vector<pair<int, int> > res;
static void add_res(int &n, int value, int c)
 {
//	printf("add %d %d\n", value, c);
	if (n == 0)
		res.push_back(make_pair(value, c)), n++;
	else if (res[n-1].first == value)
		res[n-1].second += c;
	else
		res.push_back(make_pair(value, c)), n++;
}
static bool judge(int row, int col, int i, int handled, vector<pair<int, int> > &vec, int c, int t_row)
{
	if (col == 0 && vec[i].second >= c) {
		if ((row == 0 || (handled >= c || (handled == 0 && vec[i-1].second >= c))) //up is same
			&& (row == t_row - 1 || (vec[i].second % c == 0 && (vec[i].second > c || vec[i+1].second >= c)))) // down is same
			return true;
	}
	return false;
}
static void solve(int c, int t_row, vector<pair<int, int> > &vec)
{
	int row, col, n;
	row = col = n = 0;
	for (int i = 0; i < vec.size(); i++) {
		int handled = 0;
		pair<int, int> &p = vec[i];
		int second = p.second;
		while (handled < second) {
			if (judge(row, col, i, handled, vec, c, t_row)) {
				int up = handled >= c ? p.first : i > 0 ? vec[i-1].first : p.first;
				int down = p.second >= 2 * c ? p.first : (i == vec.size() - 1 ?  p.first : vec[i+1].first);
				int value = max(abs(p.first - up), abs(p.first - down));
				add_res(n, value, c);
				handled += c;
				p.second -= handled;
				row++;
				if (p.second >= 2 * c) {
					int cl = (p.second - c) / c;
					add_res(n, 0, cl * c);
					handled += cl * c;
					p.second -= cl * c;
					row += cl;
				}
			} else {
			// oops, one by one;
				for (; col < c && handled < second; col++, handled++, vec[i].second--) {
					int value = 0;
					if (row > 0) {
						if (col < c -1 && handled < c - 1) { //up-right
							int j = i-1, k = handled;
							while ((k += vec[j].second) < c -1)
								j--;
							value = max(value, abs(p.first - vec[j].first));
						}
						if (handled < c) {
							int j = i-1, k = handled;
							while ((k += vec[j].second) < c)
								j--;
							value = max(value, abs(p.first - vec[j].first));
						}
						if (col > 0 && handled < c + 1) {
							int j = i-1, k = handled;
							while ((k += vec[j].second) < c + 1)
								j--;
							value = max(value, abs(p.first - vec[j].first));
						}
					}
					if (handled == 0 && col > 0)
						value = max(value, abs(p.first - vec[i-1].first));
					if (i < vec.size() - 1) {
						if (col < c - 1 && handled + 1 == second)
							value = max(value, abs(p.first - vec[i+1].first));
					}
					if (row < t_row - 1) {
						int left = second - handled - 1;
						if (col > 0 && left < c - 1) {
							int j = i+1, k = left;
							while ((k += vec[j].second) < c - 1)
								j++;
							value = max(value, abs(p.first - vec[j].first));
						}
						if (left < c ) {
							int j = i+1, k = left;
							while ((k += vec[j].second) < c)
								j++;
							value = max(value, abs(p.first - vec[j].first));
						}
						if (col < c - 1 && left < c + 1) {
							int j = i+1, k = left;
							while ((k += vec[j].second) < c + 1)
								j++;
							value = max(value, abs(p.first - vec[j].first));
						}
					}
					add_res(n, value, 1);
				}
				if (col == c)
					col = 0, row++;
			}
		}
		vec[i].second = second;
	}
	for (int i = 0; i < res.size(); i++)
		printf("%d %d\n", res[i].first, res[i].second);
	res.clear();
}
int main()
{
	while (true) {
		int c;
		scanf("%d", &c);
		if (!c)
			break;
		vector<pair<int, int> > vec;
		vec.reserve(1000 + 10);
		int t_row = 0;
		while (true) {
			int v, l;
			scanf("%d %d", &v, &l);
			if (!v && !l)
				break;
			vec.push_back(make_pair(v, l));
			t_row += l;
		};
		t_row /= c;
		printf("%d\n", c);
		solve(c, t_row, vec);
		printf("0 0\n");
	}
	printf("0\n");
	return 0;
}