CodeForces 289APolo the Penguin and Segments
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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
2
Input
3 7
1 2
3 3
4 7
Output
0
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;}
现在一直处于读题读不懂的状态,心累。。。
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