传送门:http://codeforces.com/contest/1047/problem/C
题意:
给定n个数,问最少要去掉几个数,使得剩下的数gcd 大于原来n个数的gcd值。
思路:
自己一开始想把每个数的因子都找出来,找到这些因子中出现次数最多且因子大于n个数的最大公约数的,(n - 次数 )就是答案。但是复杂度是1e9,差那么一点。
自己还是对素数筛理解的不够深。这道题可以枚举素数x,对于每个x,找到所有(a【i】/gcd(all)) 是x倍数的个数,就是一个次数。找这个次数的过程正好与素数筛的过程一致。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> #include <unordered_map> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ // #pragma GCC diagnostic error "-std=c++11" // #pragma comment(linker, "/stack:200000000") // #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") // #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3) #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue #define max3(a,b,c) max(max(a,b),c) typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ‘\n‘ #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 256; const double esp = 1e-8; const double PI=acos(-1.0); template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 3e5+9; const int maxnum = 1.5e7+9; int a[maxn],cnt[maxnum]; int p[maxn]; bool prime[maxnum]; int gcd(int a,int b){ if(b==0)return a; return gcd(b, a%b); } int main(){ int n,g=0;scanf("%d", &n); int ans = 0; for(int i=1; i<=n; i++){ scanf("%d", &a[i]); g = gcd(g,a[i]); } for(int i=1; i<=n; i++)cnt[a[i]/g]++; int tot = 0; for(int i=2; i<maxnum; i++){ if(!prime[i]){ int x = i;int h = 0; for(int j=i; j<maxnum; j+=i){ prime[j] = 1; h += cnt[j]; } ans = max(ans, h); } } if(!ans)puts("-1"); else printf("%d\n", n-ans); return 0; }
CF 1047 c
原文地址:https://www.cnblogs.com/ckxkexing/p/9691598.html
时间: 2024-10-10 02:35:18