Your friend has n cards.

You know that each card has a lowercase English letter on one side and a digit on the other.

Currently, your friend has laid out the cards on a table so only one side of each card is visible.

You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.

For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.

To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.

The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.

Print a single integer, the minimum number of cards you must turn over to verify your claim.

In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.

In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.

In the third sample, we need to flip the second and fourth cards.

Bob programmed a robot to navigate through a 2d maze.

The maze has some obstacles. Empty cells are denoted by the character '.', where obstacles are denoted by '#'.

There is a single robot in the maze. It's start position is denoted with the character 'S'. This position has no obstacle in it. There is also a single exit in the maze. It's position is denoted with the character 'E'. This position has no obstacle in it.

The robot can only move up, left, right, or down.

When Bob programmed the robot, he wrote down a string of digits consisting of the digits 0 to 3, inclusive. He intended for each digit to correspond to a distinct direction, and the robot would follow the directions in order to reach the exit. Unfortunately, he forgot to actually assign the directions to digits.

The robot will choose some random mapping of digits to distinct directions. The robot will map distinct digits to distinct directions. The robot will then follow the instructions according to the given string in order and chosen mapping. If an instruction would lead the robot to go off the edge of the maze or hit an obstacle, the robot will crash and break down. If the robot reaches the exit at any point, then the robot will stop following any further instructions.

Bob is having trouble debugging his robot, so he would like to determine the number of mappings of digits to directions that would lead the robot to the exit.

The first line of input will contain two integers n and m (2 ≤ n, m ≤ 50), denoting the dimensions of the maze.

The next n lines will contain exactly m characters each, denoting the maze.

Each character of the maze will be '.', '#', 'S', or 'E'.

There will be exactly one 'S' and exactly one 'E' in the maze.

The last line will contain a single string s (1 ≤ |s| ≤ 100) — the instructions given to the robot. Each character of s is a digit from 0 to 3.

Print a single integer, the number of mappings of digits to directions that will lead the robot to the exit.

For the first sample, the only valid mapping is , where D is down, L is left, U is up, R is right.

#include<bits/stdc++.h>
using namespace std;
string s[51];
int a[6],sx,sy;
int main()
{int n,m;cin>>n>>m;for(int i=0;i<n;i++)cin>>s[i];for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(s[i][j]=='S')sx=i,sy=j;string c;cin>>c;for(int i=0;i<4;i++)a[i]=i+48;int ans=0;do{int f=0;int tx=sx,ty=sy;for(int i=0;c[i];i++){if(c[i]==a[0])tx-=1;else if(c[i]==a[1])tx+=1;else if(c[i]==a[2])ty-=1;else ty+=1;if(tx<0||ty<0||tx>=n||ty>=m)break;if(s[tx][ty]=='#')break;if(s[tx][ty]=='E'){f=1;break;}}ans+=f;}while(next_permutation(a,a+4));cout<<ans;return 0;
}

6 25 5 6 8 3 12

2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613

C我逃了开根为负的情况，但是也不会是0啊，最起码是r

#include<bits/stdc++.h>
using namespace std;
int x[1005];
double y[1008];
int main()
{int n,r;cin>>n>>r;for(int i=0;i<n;i++)cin>>x[i];for(int i=0;i<n;i++){y[i]=r;for(int j=0;j<i;j++)if(fabs(x[j]-x[i])<=2.*r)y[i]=max(y[i],y[j]+sqrt(4*r*r-(x[i]-x[j])*(x[i]-x[j])));}for(int i=0;i<n;i++)printf("%.12f ",y[i]);return 0;
}

1 1 1

2

3 1 4

370000006

#include<bits/stdc++.h>
using namespace std;
const int MD=1e9+7,N=2005;
int dp[N][N];
int po(int a,int x)
{int ans=1;while(x){if(x&1) ans=1LL*ans*a%MD;a=1LL*a*a%MD;x>>=1;}return ans;
}
int main()
{int k,a,b,ta,tb,ans=0;scanf("%d%d%d",&k,&a,&b);ta=1LL*a*po(a+b,MD-2)%MD;tb=MD-ta+1;dp[1][0]=1;for(int i=1; i<=k; i++)for(int j=0; j<k-i; j++){dp[i+1][j]=(dp[i+1][j]+1LL*dp[i][j]*ta%MD)%MD;if(j+i+i>=k)ans=(ans+1LL*dp[i][j]*tb%MD*(j+i+i-k)%MD)%MD;else dp[i][j+i]=(dp[i][j+i]+1LL*dp[i][j]*tb%MD)%MD;}ans=(ans+k+1LL*a*po(b,MD-2)%MD)%MD;printf("%d\n",ans);
}