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x?∈[l,r] x \not\in [l,r]x?∈[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

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<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;
}

原文地址：https://www.cnblogs.com/zxyqzy/p/10275693.html