[link] h Write links here
[title]
Give you the number of competitors and an integer S to indicate the number of pieces per pizza.
Each parameter has three parameters, si,ai,bi;
Indicate the number of pizza pieces that the first contestant wants to eat and the increased happiness of eating the first pizza.
And eating a second pizza increases happiness.
Both pizzas can be ordered in any quantity.
But the total num has an upper limit.
S*num>=si
num is the smallest.
That means giving you a num.
Ask you how many a,b(a+b==num) you need to buy for each pizza.
Make the contestants as happy as possible.
(Participants can choose either the first or the second one at the same time)
Each parameter has three parameters, si,ai,bi;
Indicate the number of pizza pieces that the first contestant wants to eat and the increased happiness of eating the first pizza.
And eating a second pizza increases happiness.
Both pizzas can be ordered in any quantity.
But the total num has an upper limit.
S*num>=si
num is the smallest.
That means giving you a num.
Ask you how many a,b(a+b==num) you need to buy for each pizza.
Make the contestants as happy as possible.
(Participants can choose either the first or the second one at the same time)
[Abstract]
If the number of pizzas each person wants and <=S
Then you can only buy one pizza.
If you buy a or b, then everyone can only choose A or B.
Better value.
Where is the contradiction?
Although your a is big, there is no A pizza for you.
Or, it's better to buy all B pizza.
Enumerate how many B pizzas you finally bought
i=0
Pizza was eaten all the time.
i=1
Make the former S-olds of B-A eat B-pizza
i=2
Make the first 2S B-A adults eat B-pizza
Change it on the basis of i=i-1
Record who the person is and how much more that person can get from A - > B
The maximum I is num and i*S<=si
B-A is the key word in descending order?
Then you can only buy one pizza.
If you buy a or b, then everyone can only choose A or B.
Better value.
Where is the contradiction?
Although your a is big, there is no A pizza for you.
Or, it's better to buy all B pizza.
Enumerate how many B pizzas you finally bought
i=0
Pizza was eaten all the time.
i=1
Make the former S-olds of B-A eat B-pizza
i=2
Make the first 2S B-A adults eat B-pizza
Change it on the basis of i=i-1
Record who the person is and how much more that person can get from A - > B
The maximum I is num and i*S<=si
B-A is the key word in descending order?
[Number of errors]
0
[introspection]
The way to add notes is really good.
[Code]
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5;
struct abc{
ll a,b,rest,c;
};
int n;
ll S,temp,num,srest,ans;
abc a[N+10];
bool cmp(abc a,abc b)
{
return a.c > b.c;
}
int main()
{
//freopen("F:\\rush.txt","r",stdin);
scanf("%d%lld",&n,&S);
for (int i = 1;i <= n;i++){
scanf("%lld%lld%lld",&a[i].rest,&a[i].a,&a[i].b);
a[i].c = a[i].b-a[i].a;
temp += a[i].rest*a[i].a;
srest += a[i].rest;
}
//Get the maximum number of pizzas you can buy
ll l = 1,r = 1e10 + 100;
while (l <= r)
{
ll mid = (l+r)>>1;
ll temp = mid*S;
if (temp >= srest)
{
num = mid;
r = mid - 1;
}else
l = mid + 1;
}
sort(a+1,a+1+n,cmp);//In descending order B-A
ans = temp;//Start with all pizza A
//The value of this num may be very large, and you may not be able to enumerate all of it.
//If possible, try to choose greedily. As long as the number of B pizzas does not exceed num
ll temp1 = 0,tnum = 0;//tnum is the number remaining to become B better
//temp1 is the incremental value of the answers after they become B.
ll spe = num*S-srest;//Number of Pizza Pizzas Overbought
//S slice to make up a B pizza
for (int i = 1;i <= n;i++)
{
//Some of the first people turned to Pizza B.
if (a[i].c < 0)//The answer decreases when the person becomes B.
{//Use spe to try not to let him reduce - > There must have been some people from a to B, now you can use the extra filling, do not let them change.
//Premise is
if (tnum + a[i].rest + spe >= S)//That's enough.
{
ll dd = tnum + spe;
if (dd >= S)//It can be surpassed without adding.
{
ans = max(ans,temp + temp1);
}else
{
ll js = S-dd;
ans = max(ans,temp + temp1 + a[i].c*js);
}
temp1 = 0;
break;
}
}
if (tnum + a[i].rest < S)//If not more than one piece of pizza
{
temp1 += a[i].rest*a[i].c;
tnum += a[i].rest;
}else//Exceeded
{
//How many a - > b did this person choose?
temp += temp1;//Priority plus precedence
ll rr = (a[i].rest + tnum)%S;//Remainder
ll choose = a[i].rest-rr;//Then add the current one.
temp += choose*a[i].c;
tnum = rr;//The rest of the previous wheels became rr
temp1 = rr*a[i].c;//temp1 becomes rr*a[i].c
ans = max(ans,temp);//Maximum value
//Obviously B will be enough in this process.
}
}
//temp1 = 1;
//printf("%lld\n",temp1);
if (tnum + spe >= S)//The last remaining possibilities of all positive numbers can also make up an S with redundant ones.
{
ans = max(ans,temp + temp1);
}
printf("%lld\n",ans);
return 0;
}