codeforces 477B B. Dreamoon and Sets(构造)
题目链接:
Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sjfrom S, .
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
1 1
5
1 2 3 5
2 2
22
2 4 6 22
14 18 10 16
题意:
构造n个集合,每个集合里面4个数,每两个数的gcd=k,问这些数里面的最大的那个数最小是多少,和这些集合是怎样的;
思路:
1,2,3,5 7,8,9,11, 13,14,15,17 就是这样的规律,跟原来那个什么6*n+1,6*n+5折两个数有可能是质数一样,每6个相邻的数里面两两互质的就是6*n+1,6*n+2,6*n+3,6*n+5了;
然后乘上k就好了;
AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar()); for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + '0'); putchar('\n'); } const int mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e6+20; const int maxn=1e5+110; const double eps=1e-12; int main() { int n,k; read(n);read(k); cout<<(n-1)*6*k+5*k<<endl; For(i,1,n) { int t=(i-1)*6+1; printf("%d %d %d %d\n",t*k,t*k+k,t*k+2*k,t*k+4*k); } return 0; }