A:A. Nutstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You have?a?nuts and lots of boxes. The boxes have a wonderful feature: if you put?x?(x?≥?0)divisors (the spacial bars that can divide a box) to it, you get a box, divided into?x?+?1sections.

You are minimalist. Therefore, on the one hand, you are against dividing some box into more than?k?sections. On the other hand, you are against putting more than?v?nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have?b?divisors?

Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.

Input

The first line contains four space-separated integers?k,?a,?b,?v?(2?≤?k?≤?1000;?1?≤?a,?b,?v?≤?1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.

Output

Print a single integer — the answer to the problem.

Sample test(s)input3 10 3 3
output2
input3 10 1 3
output3
input100 100 1 1000
output1
Note

In the first sample you can act like this:

• Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
• Do not put any divisors into the second box. Thus, the second box has one section for the last nut.

In the end we've put all the ten nuts into boxes.

The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.

?

?花了半個多小時去理清數據的關系，開始感覺無從下手。。。我的方法是：制造相應的盒子，能放多少就盡量放到前面的盒子里面，最后統計一下就可以

#include<iostream>
#include<algorithm>
#include<math.h>
using?namespace?std;
int?aa[2000];?//盒子能放的個數
int?main()
{
????int?k,a,b,v;
????cin>>k>>a>>b>>v;
????int?ans=0;
????for?(int?i=1;i<=2000;i++)
?????????aa[i]=v;???????????????初始每個盒子開始只有一個SECTION
????for?(int?i=1;i<=2000;i++)
?????{
?????????if?(b==0)?break;
?????????if?(b+1>=k)?{aa[i]+=v*(k-1);b-=(k-1);}??有B則先用
?????????else
?????????{
?????????????aa[i]+=b*v;
?????????????break;
?????????}
?????}
????int?tem=0,ll;
????for?(int?i=1;i<=1000;i++)
????{
????????tem+=aa[i];
????????if?(tem>=a)?{ll=i;break;}
????}
????cout<<ll<<endl;
????return?0;

}?

?

B:題意很簡單，構造等差數列，求改變的數的個數最小。。

從A[1]暴力枚舉就可，不知為何我從A[N]枚舉就掛，白WA5次?#include<iostream>

#include<algorithm>
#include<math.h>
using?namespace?std;
int?a[2000],n,k;
int?main()
{
????cin>>n>>k;
????for?(int?i=1;i<=n;i++)
????cin>>a[i];
????int?ans=99999;
????int?kk;
????for?(int?i=1;i<=1000;i++)
????{
????????int?o=0;
????????int?tem=i;
????????for?(int?j=1;j<=n;j++)
????????{
????????????if?(a[j]!=tem)
????????????o++;
????????????tem+=k;
????????}
????????if?(o<ans)?{ans=o;kk=i;}
????}

????cout<<ans<<endl;
????if?(ans==0)?return?0;
????for?(int?i=1;i<=n;i++)
????{
????????if?(a[i]>kk)?{cout<<"-?"<<i<<"?"<<a[i]-kk<<endl;}
????????else?if?(a[i]<kk)?{cout<<"+?"<<i<<"?"<<kk-a[i]<<endl;}
????????kk+=k;
????}
????return?0;
}

?

C題，我是瞎搞，我的理解是使點連接的邊相對稀少，先這樣加邊：1-->2,2-->3,3-->4,n-1-->n;先見N-1條邊，

然后：1-->3,2-->4,3-->5,依次；

然后是：1--4，2-->5,....感覺這樣相對不密集。。#include<iostream>

#include<algorithm>
#include<math.h>
#include<string.h>
using?namespace?std;
int?n,p;
int?mp[1000][1000];
int?main()
{
??int?t;
??cin>>t;
??while?(t--)
??{
??????int?m=2*n+p;
??????memset(mp,0,sizeof(mp));
??????cin>>n>>p;
??????for?(int?i=1;i<n;i++)
????????mp[i][i+1]=1;
????????m=2*n+p;
????????m=m-(n-1);
????????????for?(int?i=2;i<n;i++)
????????????{
?????????????if?(m==0)?break;
?????????????for?(int?j=1;j+i<=n;j++)
?????????????{
????????????????mp[j][j+i]=1;
????????????????m--;
????????????????if?(m==0)?break;
??????????????}
????????????}
????for?(int?i=1;i<=n;i++)
????????for?(int?j=1;j<=n;j++)
????????if?(mp[i][j])
????????cout<<i<<"?"<<j<<endl;
??}
????return?0;

}

還有不得不吐槽自己的碼代碼能力，太坑了?

轉載于:https://www.cnblogs.com/forgot93/p/3604401.html

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