atcoder--------174

Problem Statement
We have
N
logs of lengths
A
1
,
A
2
,
⋯
A
N
.

We can cut these logs at most
K
times in total. When a log of length
L
is cut at a point whose distance from an end of the log is
t

(
0
<
t
<
L
)
, it becomes two logs of lengths
t
and
L
−
t
.

Find the shortest possible length of the longest log after at most
K
cuts, and print it after rounding up to an integer.

Constraints
1
≤
N
≤
2
×
10
5
0
≤
K
≤
10
9
1
≤
A
i
≤
10
9
All values in input are integers.
Input
Input is given from Standard Input in the following format:

N

K

A
1

A
2

⋯

A
N

Output
Print an integer representing the answer.

Sample Input 1
Copy
2 3
7 9
Sample Output 1
Copy
4
First, we will cut the log of length
7
at a point whose distance from an end of the log is
3.5
, resulting in two logs of length
3.5
each.
Next, we will cut the log of length
9
at a point whose distance from an end of the log is
3
, resulting in two logs of length
3
and
6
.
Lastly, we will cut the log of length
6
at a point whose distance from an end of the log is
3.3
, resulting in two logs of length
3.3
and
2.7
.
In this case, the longest length of a log will be
3.5
, which is the shortest possible result. After rounding up to an integer, the output should be
4
.

Sample Input 2
Copy
3 0
3 4 5
Sample Output 2
Copy
5
Sample Input 3
Copy
10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202
Sample Output 3
Copy
292638192

#include
using namespace std;
const int N=2e5+10;
int n,k;
int a[N];
int up(int x,int y)
{
	if (x%y==0)  return x/y-1;
	else return (x/y);
}
bool check(int len)//check函数
{
	int tot=0;
	for (int i=1;i<=n;i++)  tot+=up(a[i],len);//每次加上每个a被切的刀数！
	if (tot<=k)  return true;
	return false;
}
int main()
{
	cin>>n>>k;
	for (int i=1;i<=n;i++)  cin>>a[i];
	int l=1,r=1e9;
	while(l<r)
    {
        int mid=l+r>>1;
        if(check(mid)) r=mid;
        else l=mid+1;
    }
    cout<< r <<endl;
	return 0;
}

D - Alter Altar

Problem Statement
An altar enshrines
N
stones arranged in a row from left to right. The color of the
i
-th stone from the left
(
1
≤
i
≤
N
)
is given to you as a character
c
i
; R stands for red and W stands for white.

You can do the following two kinds of operations any number of times in any order:

Choose two stones (not necessarily adjacent) and swap them.
Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?

Constraints
2
≤
N
≤
200000
c
i
is R or W.
Input
Input is given from Standard Input in the following format:

N

c
1
c
2
.
.
.
c
N

Output
Print an integer representing the minimum number of operations needed.

Sample Input 1
Copy
4
WWRR
Sample Output 1
Copy
2
For example, the two operations below will achieve the objective.

Swap the
1
-st and
3
-rd stones from the left, resulting in RWWR.
Change the color of the
4
-th stone from the left, resulting in RWWW.
Sample Input 2
Copy
2
RR
Sample Output 2
Copy
0
It can be the case that no operation is needed.

Sample Input 3
Copy
8
WRWWRWRR
Sample Output 3
Copy
3

#include
#include
#include
#include

using namespace std;

string s;

int main()
{
    int n,cnt=0;
    cin>>n>>s;
    for(int i=0,j=n-1;i<j;i++,j--)
    {
        while(s[i]!='W' && i<j) i++;
        while(s[j]!='R' && i<j) j--;
        if(i<j) cnt++;
    }
    cout<< cnt <<endl;
    return 0;
}

C - Repsept /

Problem Statement
Takahashi loves the number
7
and multiples of
K
.

Where is the first occurrence of a multiple of
K
in the sequence
7
,
77
,
777
,
…
? (Also see Output and Sample Input/Output below.)

If the sequence contains no multiples of
K
, print -1 instead.

Constraints
1
≤
K
≤
10
6
K
is an integer.
Input
Input is given from Standard Input in the following format:

K

Output
Print an integer representing the position of the first occurrence of a multiple of
K
. (For example, if the first occurrence is the fourth element of the sequence, print 4.)

Sample Input 1
Copy
101
Sample Output 1
Copy
4
None of
7
,
77
, and
777
is a multiple of
101
, but
7777
is.

Sample Input 2
Copy
2
Sample Output 2
Copy
-1
All elements in the sequence are odd numbers; there are no multiples of
2
.

Sample Input 3
Copy
999983
Sample Output 3
Copy
999982

#include
using namespace std;
int main()
{
    int k;
    cin>>k;
    if(k%2==0 || k%5==0)
	{
        cout << -1 << endl;
        return 0;
    }
    int v=0;
    for(int i=1;;i++)
	{
        v=(v*10+7)%k;
        if(!v)
		{
            cout<< i <<endl;
            return 0;
        }
    }
    return 0;
}