[Bipartite] stove

Links: https://ac.nowcoder.com/acm/problem/21860
Source: Cattle-off network

Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 32768K, other languages 65536k
64bit the IO the Format: LLD%

Title Description

J small start working, trying to earn money to buy a gas stove.
The first day, a small wage of n j yuan, a day after his salary more than the day before d yuan.
Known gas stove needs m yuan, seeking small j for at least a few days to buy gas stove.

Enter a description:

Four integers n, m, d, x 
are respectively the first day of the small pay j, Gas prices, increase the amount of salary per day, the answer is no more than x

Output Description:

A number indicates the answer

Example 1

Entry

copy

10 100 20 100

Export

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4

Explanation

10+30+50+70>=100

Remarks:

0 ≦ n &, d≤1e9, the n-+ d> 0 
1≤m≤1e18 
1≤x≤1e9 

meaning of the questions: can get paid every day, at the beginning is n, every day may be the day before the overpaid d (note d may be zero), and asked how many days the minimum sum you receive salary can exceed m yuan to buy gas stove need, no more than x number of days 
thinking: summation problem, and more are likely to increase the value of d day, so this is a beginning is n, the number of arithmetic d tolerance column summing arithmetic sequence summation formula: Sn = a0 * n + ( n * (n-1) * d / 2), m and the maximum is noted 1e18, to let arithmetic sequence sum Sn> = m, the Sn would exceed the range of a long long, so to formula deformation, 
n is an initial value, in the number of days, d is the tolerance, m is at least the sum of the required value larger than, get

　　　　n*in+(in*(in-1)*d)/2>=m
　　　　2*(n*in)+(in(in-1)*d)>=2*m
　　　　in*(in-1)*d>=2*(m-(n*in))

While noting that 1 <= in <= x, so I think half of what the answer

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll n,m,d,x;
 5 bool check(ll in){
 6     return in*(in-1)*d>=2*(m-(n*in));
 7 }
 8 int main(){
 9     cin>>n>>m>>d>>x;
10     ll l=1,r=x,mid=(l+r)>>1;
11     while(l<r){
12         if(check(mid))r=mid;
13         else l=mid+1;
14         mid=(l+r)>>1;
15     }
16     printf("%lld\n",mid);
17 }
18 /***
19 n*in+(in*(in-1)*d)/2>=m
20 2*(n*in)+(in(in-1)*d)>=2*m
21 in*(in-1)*d>=2*(m-(n*in))
22 ***/