Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2)

Problem A

Between the offices

Water problem, water a water.

 
 
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 char s[200];
 5 int main()
 6 {
 7     cin>>n;
 8     int cnt1=0,cnt2=0;
 9     scanf("%s",s);
10     for(int i=0;i<n-1;i++)
11     {
12         if(s[i]==‘S‘ && s[i+1]==‘F‘) cnt1++;
13         else if(s[i]==‘F‘ && s[i+1]==‘S‘) cnt2++;
14     }
15     if(cnt1>cnt2) puts("YES");
16     else puts("NO");
17     return 0;
18 }

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Problem B

Save the problem!

give you the number n, let you give a sum of money m, and the type of money N, ask for these kinds of money

The number of species that make up M is N.

Idea: I was constructed with 5 and 22 kinds of money, we consider the composition of 10 is only two kinds, one is 2 2 2 2 2, the other is

5 5, then 20 is three kinds, 30 is four, 40 is five, has been so constructed, forgot to consider a situation WA once ...

 
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 int main()
 5 {
 6     scanf("%d",&n);
 7     if(n==1)
 8     {
 9         puts("1 1");
10         puts("1");
11         return 0;
12     }
13     printf("%d 2\n",(n-1)*10);
14     puts("5 2");
15     return 0;
16 }

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Promblem C

Ordering Pizza

There are now n people need to eat si slice pizza, a total of two kinds of pizza A, a, I personally eat a slice a pizza obtained

The pleasure value is ai,b for bi, a slice of pizza can be divided into s slices, we need to buy the fewest pizzas and ask you what the maximum pleasure value is.

Idea: Let's get everyone to eat a pizza with a good value and then count the number of slices of a pizza and the number of slices in B pizza,

A and B to the S to take the film, the rest is uncertain, if a+b>s buy two kinds of pizza each one, if a+b<=s

Then buy a pizza, both of which are simulated to fetch a large number of pleasant values.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+5;
ll _abs(ll x)
{
    if(x>=0) return x;
    else return -x;
}
struct node
{
    ll s,a,b;
    bool operator < (const node &rhy)const
    {
        return _abs(a-b)<_abs(rhy.a-rhy.b);
    }
}p[N];
ll n,S;
ll work(ll c1,ll c2)
{
    ll ans=0;
    if(c1)
    {
        for(int i=0;i<n;i++)
        {
            if(p[i].a<p[i].b) continue;
            if(p[i].s>=c1) ans+=c1*(p[i].a-p[i].b);
            else ans+=p[i].s*(p[i].a-p[i].b);
            c1-=p[i].s;
            if(c1<=0) break;
        }
    }
    else
    {
        for(int i=0;i<n;i++)
        {
            if(p[i].a>=p[i].b) continue;
            if(p[i].s>=c2) ans+=c2*(p[i].b-p[i].a);
            else ans+=p[i].s*(p[i].b-p[i].a);
            c2-=p[i].s;
            if(c2<=0) break;
        }
    }
    return ans;
}
int main()
{
    cin>>n>>S;
    ll sum=0,ans=0;
    ll res1=0,res2=0;
    for(int i=0;i<n;i++)
    {
        scanf("%I64d%I64d%I64d",&p[i].s,&p[i].a,&p[i].b);
        ans+=p[i].s*max(p[i].a,p[i].b);
        if(p[i].a>=p[i].b) res1+=p[i].s;
        else res2+=p[i].s;
    }
    res1=res1%S; res2=res2%S;
    if(res1+res2>S)
    {
        printf("%I64d\n",ans);
        return 0;
    }
    ll res=1e18;
    sort(p,p+n);
    res=min(res,work(0,res2));
    res=min(res,work(res1,0));
    printf("%I64d\n",ans-res);
    return 0;
}

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Problem E

Buy Low Sell High

To give you some stock price per day, every day you can choose to buy, sell, or do nothing, ask you

The maximum amount of profit.

Idea: puzzles, I will not! To maintain the minimum value with the priority queue, we take a number B, if this number is higher than the heap top

Element A is small, then B is added to the priority queue, if B is larger than a, delete a, add two B, (b-a) join ans,

The first B is the equivalent of turning a into B and the second b being itself.

 
 
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=3*1e5+5;
 4 int n,a[N];
 5 int main()
 6 {
 7     scanf("%d",&n);
 8     for(int i=1;i<=n;i++) scanf("%d",&a[i]);
 9     priority_queue<int,vector<int>,greater<int> > Q;
10     long long ans=0;
11     for(int i=1;i<=n;i++)
12     {
13         if(Q.empty())
14         {
15             Q.push(a[i]);
16             continue;
17         }
18         int t=Q.top();
19         if(t>=a[i]) Q.push(a[i]);
20         else
21         {
22             ans+=a[i]-t;
23             Q.pop();
24             Q.push(a[i]); Q.push(a[i]);
25         }
26     }
27     printf("%I64d\n",ans);
28     return 0;
29 }

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Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2)