题意翻译

题意简述

给出CF1101A Minimum Integer 模拟_#define

输入格式

第一行一个正整数CF1101A Minimum Integer 模拟_#include_02

接下来CF1101A Minimum Integer 模拟_#include_03

输出格式

对于每一组询问输出一行表示答案

题目描述

You are given CF1101A Minimum Integer 模拟_#include_04

Given three integers CF1101A Minimum Integer 模拟_git_05

Can you answer all the queries?

Recall that a number CF1101A Minimum Integer 模拟_git_06

输入输出格式

输入格式：

The first line contains one integer CF1101A Minimum Integer 模拟_#define_07

Then CF1101A Minimum Integer 模拟_#define_08

输出格式：

For each query print one integer: the answer to this query.

输入输出样例

输入样例#1：

复制

5
2 4 2
5 10 4
3 10 1
1 2 3
4 6 5

输出样例#1： 复制

6
4
1
3
10
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 100005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int q;
ll l, r, d;

int main() {
    ios::sync_with_stdio(0);
    cin >> q;
    while (q--) {
        cin >> l >> r >> d;
        ll L, R;
        if (l%d != 0) {
            L = (l / d);
        }
        else if (l%d == 0)L = l / d - 1;
        if (r%d == 0)R = r / d + 1;
        else if (r%d != 0)R = r / d + 1;
        if (L == 0) {
            cout << d * R << endl;
        }
        else {
            cout << 1 * d << endl;
        }
    }
    return 0;
}

 

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