codeforces !Hasan
Mixed Dimensions Onsite Round will host N contestants this year.
Moath and Saif have been preparing the contest hall, Moath needs X minutes to set up a computer for the contest, while Saif needs Yminutes to set up a computer. Each one of them works separately on one computer at a time.
Hasan is concerned they won't finish setting up the PCs in time, can you help Hasan figure out the minimum time required by Moath and Saif to set up all N PCs for the contest?
Input
The input contains 3 integers, N, X, and Y (1 ≤ N, X, Y ≤ 109), the number of PCs to set up, the amount of minutes Moath needs to set up a computer, and the amount of minutes Saif needs to set up a computer, respectively.
Output
On a single line, print one integer, the minimum number of minutes required by Moath and Saif to set up all the PCs for the contest.
Examples
input
5 3 4
output
9
input
100 10 1
output
91
———————————————————————————————————————————————————————————— 说一说今天的雪崩经历……这道题是十分的令人崩溃啊，wa了23发终于过了好吧，堪称励志。
就是说有n个电脑，甲能a分钟修1个，乙能b分钟修一个，问最少多少时间修完
先拿到题看，嗯，容易啊，就是对于，反正ab一直在做，瞎搞搞就好了么，然后队友就给了我一个公式：n*a*b/(a+b)，结果可想而知，爆int了。
然后发现了问题，改成longlong，然后就发现了最严重的问题……这个结果，double装不下……不下……下……
瞬间懵逼。
当时学弟还在车上，然后过来以后讲了下方法，事实上已经快要正确了，结果自己加了个特判……然后挖了个坑自己跳进去了。
悲剧啊……
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嘛，这道题两种解法，一种是我们假设这个行为在某个时间结束，那么结束的时间总是固定的吧，然后我们假设在这个结束时间甲修了x个，那么乙就修了n-x，那公式是这样的：a*x=b(n-x)，然后把这个方程式解出来，x可能是个小数，因为甲修完x个的时候可能乙没东西修了，但是从公式上来说乙是要修到一半的。
那么有了这个小数，问题就变成了最后一个东西是甲修还是乙修，讨论一下就好了
第二种解法比较计算机思维，就是既然放不下，那我也不放了，直接用二分搜索查找答案，250ms，log2 (1023)的复杂度也不过23/log102,轻松愉快的就找到了目标，当然过程得讨论一下，就是如果当前的时间是够的，那么rig=mid,如果当前不够，就lef=mid+1,这样最终会变成lef==rig的情况，然后直接输出就好啦。
——————————————————————code version 1——————————————————————————————
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
#include<vector>
#define V 100+10
#define E 10000+10
#define ll long long
const int INF=0x3f3f3f3f;
using namespace std;
ll mina=INF,shi;
int main()
{
ll n;
ll a,b,c,d;
ll x,y;
cin>>n>>a>>b;
c=min(a,b);d=max(a,b);
/*if(d==c){
if(n%2==0)cout<<n*c/2<<endl;
else cout<<(n/2+1)*c<<endl;
}*/
y=d*n/(c+d)+1;
if(y>n)cout<<c*n<<endl;
else {
for(x=y;x>=0;x--){
if(c*x<d*(n-x)){
mina=min(c*(x+1),d*(n-x));
break;
}
}
cout<<mina<<endl;
}
}————————————————————version 2—————————————————————————————————
#include<stdio.h> #include<string.h> #include<iostream> #include<vector> #include<algorithm> #include<string> #include<queue> #include<climits> #include<map> #include<stack> #include<list> #define file_in freopen("input.txt","r",stdin) #define MAX 8000 #define HASH 100019 using namespace std; #define ll long long #define FF(x,y) for(int i=x;i<y;i++) int main() { ll n, a, b; cin >> n >> a >> b; ll lef = 1; ll rig = 1e23; while (1) { ll mid = (lef + rig) >> 1; if (mid / a + mid / b >= n){ rig = mid; } else { lef = mid+1; } if (lef >= rig) { cout << lef << endl; return 0; } } }
