Codeforces-289a I Polo the Penguin and Segments

A. Polo the Penguin and Segments

time limit per test
2 seconds

memory limit per test
256 megabytes

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 test(s)

Input

2 3
1 2
3 4

Output

2

Input

3 7
1 23 3
4 7

Output

0

——————————————————计划满满的分割线——————————————————

思路：此题考的是读懂题意。。。。。。读懂后，你发现它其实就是让你统计整数的个数，看它能否被k整除，不能的话，需要增加ans或者减少ans才可以（ans取最小的）

代码如下：

#include <stdio.h>
int main(){
int n, add = 0, k;
int x, y, mod, ans;
scanf("%d%d", &n, &k);
while(n--){
scanf("%d%d", &x, &y);
add += (y - x + 1);
}
mod = add % k;
if(mod) ans = k - mod;
else ans = 0;
printf("%d\n", ans);
return 0;
}