CF1101A Minimum Integer 模拟
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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 otin [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
5 2 4 2 5 10 4 3 10 1 1 2 3 4 6 5
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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