codeforces-749【思维】

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A. Bachgold Problem

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum
possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and
has exactly two positive integer divisors — 1 and k.

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n.
You can print them in any order. If there are several optimal solution, print any of them.

Examples

input

5

output

2
2 3

input

6

output

3
2 2 2

大意：任意一个数 n（2~100000） 拆成 k 个素数之和，要求尽可能使得 k 最大

思路：n 要么为奇数（拆成 ( n/2 ) 个2 和 1 ，转化一下就是 ( n/2-1 ) 个 2 和 3 ），要么为偶数（拆成 n/2 个2 ）

#include<cstdio>
#include<cmath>
using namespace std;
int n;
int main()
{
while(~scanf("%d",&n))
{
printf("%d\n",n/2);
if(n&1)
{
for(int i=1;i<n/2;i++)
printf("2 ");
puts("3");
}
else
{
for(int i=1;i<=n/2;i++)
printf(i==1?"2":" 2");
puts("");
}
}
return 0;
}

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B. Parallelogram is Back

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Long time ago Alex created an interesting problem about parallelogram. The input data for this problem contained four integer points on the Cartesian plane, that defined the set of vertices of some non-degenerate (positive area) parallelogram. Points not necessary
were given in the order of clockwise or counterclockwise traversal.

Alex had very nice test for this problem, but is somehow happened that the last line of the input was lost and now he has only three out of four points of the original parallelogram. He remembers that test was so good that he asks you to restore it given only
these three points.

Input

The input consists of three lines, each containing a pair of integer coordinates xi and yi ( - 1000 ≤ xi, yi ≤ 1000).
It's guaranteed that these three points do not lie on the same line and no two of them coincide.

Output

First print integer k — the number of ways to add one new integer point such that the obtained set defines some parallelogram of positive
area. There is no requirement for the points to be arranged in any special order (like traversal), they just define the set of vertices.

Then print k lines, each containing a pair of integer — possible coordinates of the fourth point.

Example

input

0 0
1 0
0 1

output

3
1 -1
-1 1
1 1

Note

If you need clarification of what parallelogram is, please check Wikipedia page:
https://en.wikipedia.org/wiki/Parallelogram
大意：给出平行四边形的三个点，求第四个点的可能数目，并输出这个点。

以为是 B 题不会这么水，敲完之后不敢交。又回去读了好几遍题，想去找坑，实在是没找到。zz 浪费十多分钟

#include<cstdio>
#include<algorithm>
using namespace std;
int x1,x2,x3,y1,y2,y3;
int main()
{
while(~scanf("%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3))
{
puts("3");
printf("%d %d\n",x1+x2-x3,y1+y2-y3);
printf("%d %d\n",x1+x3-x2,y1+y3-y2);
printf("%d %d\n",x2+x3-x1,y2+y3-y1);
}
return 0;
}

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C. Voting

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n employees in Alternative Cake Manufacturing (ACM). They are now voting on some very important question and the leading
world media are trying to predict the outcome of the vote.

Each of the employees belongs to one of two fractions: depublicans or remocrats, and these two fractions have opposite opinions on what should be the outcome of the vote. The voting procedure is rather complicated:

Each of n employees makes a statement. They make statements one by one starting from employees 1 and
finishing with employee n. If at the moment when it's time for the i-th
employee to make a statement he no longer has the right to vote, he just skips his turn (and no longer takes part in this voting).

When employee makes a statement, he can do nothing or declare that one of the other employees no longer has a right to vote. It's allowed to deny from voting people who already made the statement or people who are only waiting to do so. If someone is denied
from voting he no longer participates in the voting till the very end.

When all employees are done with their statements, the procedure repeats: again, each employees starting from 1 and finishing withn who
are still eligible to vote make their statements.

The process repeats until there is only one employee eligible to vote remaining and he determines the outcome of the whole voting. Of course, he votes for the decision suitable for his fraction.

You know the order employees are going to vote and that they behave optimal (and they also know the order and who belongs to which fraction). Predict the outcome of the vote.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) —
the number of employees.

The next line contains n characters. The i-th
character is 'D' if the i-th
employee is from depublicans fraction or 'R' if he is from remocrats.

Output

Print 'D' if the outcome of the vote will be suitable for depublicans and 'R'
if remocrats will win.

Examples

input

5DDRRR

output

D

input

6DDRRRR

output

R

Note

Consider one of the voting scenarios for the first sample:

Employee 1 denies employee 5 to
vote.

Employee 2 denies employee 3 to
vote.

Employee 3 has no right to vote and skips his turn (he was denied by employee 2).

Employee 4 denies employee 2 to
vote.

Employee 5 has no right to vote and skips his turn (he was denied by employee 1).

Employee 1 denies employee 4.

Only employee 1 now has the right to vote so the voting ends with the victory of depublicans.

大意：有两个政党，n 个人进行投票的结果。每个人都知道其他人的投票结果和顺序，而且都会选择最优的方式（就是本着让自己支持政党获胜的想法）否定他人。现在进行一种操作：每个人可以否定另一个人的投票（前提是他自己没有被否定），投票者被否定之后此前的投票无效，直到最后剩余一个人为止。输出最后这个人所支持的政党。

思路：用 queue 或 set 维护投票的人即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int n;
char str[200010];
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",str+1);
queue<int> QD;
queue<int> QR;
for(int i=1;i<=n;i++)
{
if(str[i]=='D')
QD.push(i);
else
QR.push(i);
}
while(QD.size()&&QR.size())
{
int x=QD.front();
QD.pop();
int y=QR.front();
QR.pop();
if(x<y)
QD.push(x+n);
else
QR.push(y+n);
}
if(!QD.empty())
puts("D");
else
puts("R");
}
return 0;
}