Codeforces Round #274 (Div. 2) B. Towers

As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists
of ai cubes
stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example,
if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).

The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would
never put the cube on the same tower from which it was removed because he thinks it's a waste of time.

Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him
accomplish this task.

Input

The first line contains two space-separated positive integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000)
— the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1 ≤ ai ≤ 104)
— the towers' initial heights.

Output

In the first line print two space-separated non-negative integers s and m (m ≤ k).
The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number
of operations needed for that.

In the next m lines print the description of each operation as two positive integers i and j,
each of them lies within limits from 1 to n. They represent
that Petya took the top cube from the i-th tower and put in on the j-th
one (i ≠ j). Note that in the process of performing operations the heights of some towers can become equal to zero.

If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.

Sample test(s)

input

3 2
5 8 5

output

0 2
2 1
2 3

input

3 4
2 2 4

output

1 1
3 2

input

5 3
8 3 2 6 3

output

3 3
1 3
1 2
1 3

题意：给出n个塔，由a[i]个砖块堆起来。整个体系的稳定系数为：最高的高度减去最矮的高度。

能够移动k次，求移动k次中，能够得到的最低的稳定系数，以及怎样移动，每次仅仅能移动1个方块。

思路：模拟

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;

struct Node {
	int x, id;
} node[maxn];
struct Op {
	int x, y;
} q[maxn];
int ans[maxn];

bool cmp(Node a, Node b) {
	return a.x < b.x;
}

int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &node[i].x);
		node[i].id = i;
	}
	sort(node+1, node+1+n, cmp);
	ans[0] = node[n].x - node[1].x;
	for (int i = 1; i <= k; i++) {
		sort(node+1, node+1+n, cmp);
		node[1].x++, node[n].x--;
		q[i].y = node[1].id, q[i].x = node[n].id;
		sort(node+1, node+1+n, cmp);
		ans[i] = node[n].x - node[1].x;
	}
	int pos = 0, Max = ans[0];
	for (int i = 1; i <= k; i++)
		if (ans[i] < Max)
			Max = ans[i], pos = i;
	printf("%d %d\n", Max, pos);
	for (int i = 1; i <= pos; i++)
		printf("%d %d\n", q[i].x, q[i].y);
	return 0;
}

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