CF1101A Minimum Integer 模拟

题意翻译

题意简述

给出qqq组询问，每组询问给出l,r,dl,r,dl,r,d，求一个最小的正整数xxx满足d∣x d | x\ d∣x 且x̸∈[l,r] x \not\in [l,r]x̸∈[l,r]

输入格式

第一行一个正整数q(1≤q≤500)q(1 \leq q \leq 500)q(1≤q≤500)

接下来qqq行每行三个正整数l,r,d(1≤l≤r≤109,1≤d≤109)l,r,d(1 \leq l \leq r \leq 10^9 , 1 \leq d \leq 10^9)l,r,d(1≤l≤r≤109,1≤d≤109)表示一组询问

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 q q q queries in the following form:

Given three integers li l_i li , ri r_i ri and di d_i di , find minimum positive integer xi x_i xi such that it is divisible by di d_i di and it does not belong to the segment [li,ri] [l_i, r_i] [li,ri] .

Can you answer all the queries?

Recall that a number x x x belongs to segment [l,r] [l, r] [l,r] if l≤x≤r l \le x \le r l≤x≤r .

输入输出格式

输入格式：

The first line contains one integer q q q ( 1≤q≤500 1 \le q \le 500 1≤q≤500 ) — the number of queries.

Then q q q lines follow, each containing a query given in the format li l_i li ri r_i ri di d_i di ( 1≤li≤ri≤109 1 \le l_i \le r_i \le 10^9 1≤li≤ri≤109 , 1≤di≤109 1 \le d_i \le 10^9 1≤di≤109 ). li l_i li , ri r_i ri and di d_i di are integers.

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

输出样例#1： 复制

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<string>

#include<cmath>

#include<map>

#include<set>

#include<vector>

#include<queue>

#include<bitset>

#include<ctime>

#include<deque>

#include<stack>

#include<functional>

#include<sstream>

//#include<cctype>

//#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<int, int> pii;

#define pi acos(-1.0)

//const int N = 1005;

#define REP(i,n) for(int i=0;i<(n);i++)

typedef pair<int, int> 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;

}