传送门
文章目录
- 题意:
- 思路:
题意:
有两种物品分别有x,yx,yx,y个,每次可以从一个拿出aaa个,另一个拿出bbb个分成一组,问最多能分成多少组。
思路:
这个题有一个显然的单调性,所以二分一个midmidmid表示拿了midmidmid个组,考虑如何checkcheckcheck。
设kkk为从第一个物品中拿了kkk次aaa个,那么可以列出如下两个不等式:x≤k∗a+(mid−k)∗bx\le k*a+(mid-k)*bx≤k∗a+(mid−k)∗by≤(mid−k)∗a+k∗by\le (mid-k)*a+k*by≤(mid−k)∗a+k∗b
将变量kkk单独拿出来:
k≥y−mid∗ab−ak\ge \frac{y-mid*a}{b-a}k≥b−ay−mid∗a
k≤x−mid∗ba−bk\le \frac{x-mid*b}{a-b}k≤a−bx−mid∗b
当然kkk还有一个自然范围[0,mid][0,mid][0,mid],所以问题转换成了是否存在这样的left≤k≤rightleft \le k \le rightleft≤k≤right,由于是浮点数,左边上取整,右边下取整即可。
// Problem: G. Gift Set
// Contest: Codeforces - Codeforces Round #725 (Div. 3)
// URL: https://codeforces.com/contest/1538/problem/G
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int a,b,x,y;bool check(LL mid) {LL left=ceil(1.0l*(y-a*mid)/(b-a));LL right=floor(1.0l*(x-b*mid)/(a-b));if(left<=mid&&right>=0&&left<=right) return true;return false;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d%d%d%d",&x,&y,&a,&b);if(a==b) {printf("%d\n",min(x,y)/a);continue;}if(a<b) swap(a,b);int l=0,r=1e9,ans=-1;while(l<=r) {int mid=l+r>>1;if(check(mid)) ans=mid,l=mid+1;else r=mid-1;}printf("%d\n",ans);}return 0;
}
/**/