題意翻譯

題意簡述

給出qqq組詢問，每組詢問給出l,r,dl,r,dl,r,d，求一個最小的正整數xxx滿足d∣x?d | x\ d∣x?且x?∈[l,r] x \not\in [l,r]x?∈[l,r]

輸入格式

第一行一個正整數q(1≤q≤500)q(1 \leq q \leq 500)q(1≤q≤500)

接下來qqq行每行三個正整數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)表示一組詢問

輸出格式

對于每一組詢問輸出一行表示答案

題目描述

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 .

輸入輸出格式

輸入格式：

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.

輸出格式：

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

輸入輸出樣例

輸入樣例#1： 復制 5 2 4 2 5 10 4 3 10 1 1 2 3 4 6 5 輸出樣例#1： 復制 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; }

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轉載于:https://www.cnblogs.com/zxyqzy/p/10275693.html

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