CodeForces 289A Polo the Penguin and Segments

最新推荐文章于 2017-08-21 20:36:21 发布

Coder_Jeremy Wu

最新推荐文章于 2017-08-21 20:36:21 发布

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分类专栏： Codeforces 文章标签： codeforces

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这是一道关于小企鹅Polo和他的区间问题的编程题。题目要求找到使一组不相交区间长度和能被k整除的最小操作次数，每次操作可以将任意区间的左右端点增加或减少1。

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Polo the Penguin and Segments

Description

Little penguin Polo adores integer segments, that is, pairs of integers [l; r](l ≤ r).

He has a set that consists of n integer segments: [l1; r1], [l2; r2], …, [ln; rn]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [l; r] to either segment [l - 1; r], or to segment [l; r + 1].

The value of a set of segments that consists of n segments [l1; r1], [l2; r2], …, [ln; rn] is the number of integers x, such that there is integer j, for which the following inequality holds, lj ≤ x ≤ rj.

Find the minimum number of moves needed to make the value of the set of Polo’s segments divisible by k.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 105). Each of the following n lines contain a segment as a pair of integers li and ri ( - 105 ≤ li ≤ ri ≤ 105), separated by a space.

It is guaranteed that no two segments intersect. In other words, for any two integers i, j(1 ≤ i < j ≤ n) the following inequality holds, min(ri, rj) < max(li, lj).

Output

In a single line print a single integer — the answer to the problem.

Sample Input

Input

2 3
1 2
3 4

Output

Input

3 7
1 2
3 3
4 7

Output

我不得不说，恶心死了这样的题目
解释一大堆，绕来绕去，全是废话，还看不懂题意
没做出来，气死我了，看了半天题意没搞懂
百度也百度不到详细题意
所以我在这里详细的讲解一下

题意：

输入n行，每行两个元素，表示一个区间，现在这n个区间每次只能在左端减1或者在右端加1，问如果这n个区间的长度和能被k整除，需要进行多少次变化？输出结果,
英文题目看不懂什么意思，废话一堆

AC

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    int sum=0;
    int l,r;
    while(n--)
    {
        scanf("%d%d",&l,&r);
        sum+=(r-l+1);//区间的长度就是区间内元素的个数
    }
    sum=(k-sum%k)%k;//这里多一次求余k的目的就是为了防止出现集合正好被整除的特殊情况！
    printf("%d",sum);
    return 0;
}