You are given qq queries in the following form:
Given three integers lili, riri and didi, find minimum positive integer xixi such that it is divisible by didi and it does not belong to the segment [li,ri][li,ri].
Can you answer all the queries?
Recall that a number xx belongs to segment [l,r][l,r] if l≤x≤rl≤x≤r.
Input
The first line contains one integer qq (1≤q≤5001≤q≤500) — the number of queries.
Then qq lines follow, each containing a query given in the format lili riri didi (1≤li≤ri≤1091≤li≤ri≤109, 1≤di≤1091≤di≤109). lili, riri and didi are integers.
Output
For each query print one integer: the answer to this query.
Example
Input
5
2 4 2
5 10 4
3 10 1
1 2 3
4 6 5
Output
6
4
1
3
10 题目链接：CodeForces - 1101A水题一个，但是数据量略大不足以让我们暴力随便过。
那么便思考一下找公式就行了，
观察可知，当d小于L的时候，答案就是d
否则，答案是(r/d+1)*d; 我的AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int q;
int l,r;
int d;
int main()
{
gbtb;
cin>>q;
while(q--)
{
cin>>l>>r>>d;
int flag=;
if(d<l)
{
cout<<d<<endl;
continue;
}else
{
int ans=(r/d+)*d;
cout<<ans<<endl;
}
}
return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}