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Codeforces 346C Number Transformation II 构造

时间:2016-04-23 19:43:02      阅读:152      评论:0      收藏:0      [点我收藏+]

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题目链接:点击打开链接

= = 990+ms卡过

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define N 100010
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define ll int
ll n,m,k,a,b;
ll x[N];
bool cmp(ll z,ll y){return z>y;}
set<int>myset;
set<int>::iterator p;
int main(){
	ll i,j;
	while(cin>>n){
		myset.clear();
		for(i=1;i<=n;i++)scanf("%d",&j),myset.insert(j);
		cin>>a>>b;
		if(a==b){puts("0");continue;}
		n = myset.size();
		i = 1;
		for(p = myset.begin(); p!=myset.end(); p++,i++)x[i] = *p;
		sort(x+1,x+1+n,cmp);
		ll step = 0;
		ll l = 1;
		while(x[l]>a && l<=n)l++;
		while(x[l]>(a>>1) && (a-x[l])>(a-b))l++;
		if(l>n){cout<<a-b<<endl;continue;}

		while(a!=b){
			step++;
			ll now = 1, cha = a-b;

			for(i = l;i<=n;i++)
			{
				if(now>=x[i])break;
				ll tmp = a-((a/x[i])*x[i]);
				if(tmp<=now || tmp>cha)continue;
				now = tmp;
			}
			a-=now;
			while(x[l]>a&&l<=n)l++;
			while(x[l]>(a>>1) && (a-x[l])>cha)l++;
			if(l>n)break;
		}
		cout<<step+a-b<<endl;
	}
	return 0;
}


Codeforces 346C Number Transformation II 构造

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原文地址:http://www.cnblogs.com/yxwkf/p/5425148.html

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