Codeforces-289a I Polo the Penguin and Segments
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.
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).
In a single line print a single integer — the answer to the problem.
2 3 1 2 3 4
2
3 7 1 2 3 3 4 7
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
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;
}