My solution is depend on next_permutation in C++.

I got AC with O(N*N);

Idea: use mathematical induction and prove that solution exist always.
Induction gives a solution.

this is my solution
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
   long long a,b,s=0,d=0,k=0,jama,kk=0,ss=0,dd=0;
   cin>>b;
   a=b;
   b=b-1;
   while(b-1>0)
   { jama=b%10;
       k++;
       if(k<=3)
       {
           s+=jama;
       }
       else{
        d+=jama;
       }
       b/=10;

   }

   int i=0;
   if(s-d==0 )
   {
      i++;

   }

   a=a+1;
   int joma;
    while(a>0)
   { joma=a%10;
       kk++;
       if(kk<=3)
       {
           ss+=joma;
       }
       else{
        dd+=joma;
       }
       a/=10;

   }
   if(ss-dd==0 )
   {
      i++;

   }
   if(i>0)
   {
       cout<< "Yes";

   }
   else
    cout<< "No";
    return 0;
}

#include <iostream>
using namespace std;


int main()
{
    int k = 0;
    long long int s = 0;
    char num[52];
    int arr[51];
    bool flag = true;
    cin >> num;
    for(int i = 0; flag; i++)
    {
        switch(num[i])
        {
            case '0':
            {
                arr[i] = 0;
                k++;
                break;
            }
            case '1':
            {
                arr[i] = 1;
                k++;
                break;
            }
            case '2':
            {
                arr[i] = 2;
                k++;
                break;
            }
            case '3':
            {
                arr[i] = 3;
                k++;
                break;
            }
            case '4':
            {
                arr[i] = 4;
                k++;
                break;
            }
            case '5':
            {
                arr[i] = 5;
                k++;
                break;
            }
            case '6':
            {
                arr[i] = 6;
                k++;
                break;
            }
            case '7':
            {
                arr[i] = 7;
                k++;
                break;
            }
            case '8':
            {
                arr[i] = 8;
                k++;
                break;
            }
            case '9':
            {
                arr[i] = 9;
                k++;
                break;
            }
            case '\0':
            {
                flag = false;
                break;
            }
        }
    }
    for(int i = k-1; i+1 > 6;)
    {
        s = s + arr[i];
        s = s + arr[i-1]*3;
        s = s + arr[i-2]*2;
        s = s + arr[i-3]*6;
        s = s + arr[i-4]*4;
        s = s + arr[i-5]*5;
        i = i - 6;
        k = k - 6;
    }
    if(k > 0)
    {
        s = s + arr[k-1];
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*3;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*2;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*6;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*4;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*5;
        k--;
    }
    cout << s%7;
    return 0;
}

Edited by author 16.11.2013 14:57

Edited by author 16.11.2013 14:57

import java.util.Scanner;


public class A380 {
    public static void main(String agrs[]){
        Scanner sys=new Scanner(System.in);
        String input=sys.next();
        String[] part=input.split("(?<=\\d)(?=\\D)");
        int num=Integer.parseInt(part[0]);
        if(num==1 || num==2){
            if(part[1].equals("A")||part[1].equals("D"))
                System.out.println("window");
            else if(part[1].equals("B")||part[1].equals("C"))
                System.out.println("aisle");

        }
        else if(num>=3 && num<=20){
            if(part[1].equals("A")||part[1].equals("F"))
                System.out.println("window");
            else if(part[1].equals("B")||part[1].equals("C") || part[1].equals("D")||part[1].equals("E"))
                System.out.println("aisle");

        }
        else
            if(num>=21 && num<=65){
            if(part[1].equals("A")||part[1].equals("K"))
                System.out.println("window");
            else if(part[1].equals("C")||part[1].equals("D") || part[1].equals("H")||part[1].equals("G"))
                System.out.println("aisle");
            else if(part[1].equals("B")||part[1].equals("E") || part[1].equals("F")||part[1].equals("J"))
                System.out.println("niether");
        }

    }

}

Time limit is 1.0s, my program executes test 11 in 0.187s the result is TLE, why ?

I can't pass test 4, I just can't,,,,,


thanks,, problem founded....

Edited by author 29.11.2013 11:03

#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
if (39+2*a>a+2*b) cout << 39+2*a;
else;cout << 40+2*b;
return 0;
}

Enjoy:

var n:1..1000; i,e,l,m:1..1000; s:string[1]; begin readln(n); for i:=1 to n do begin readln(s); if s[1]='E' then inc(e); if s[1]='L' then inc(l); if s[1]='M' then inc(m); end; if (e>=l) and (e>=m) then writeln('Emperor Penguin'); if (l>=e) and (l>=m) then writeln('Little Penguin'); if (m>=e) and (m>=l) then writeln('Macaroni Penguin'); end.

Good luck!

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int br[10024],i,m;
    double n,a;
    float sum;
    cin>>n>>a;
    sum=100/a;
    for(i=1;i<=10024;i++)
    {
        br[i]=0;
    }
    for(i=1;i<=a;i++)
    {
        cin>>m;
        br[m]++;
    }
    for(i=1;i<=n;i++)
    {

         printf("%.2lf",(br[i]*sum));
         cout<<"%"<<endl;
    }
    return 0;
}

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <stack>
#include <algorithm>
#include <bitset>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <iomanip>

#define F first
#define S second
#define ll long long
#define len length()
#define sqr(x) x*x
#define pb push_back
#define mp make_pair
#define sz(x) ((int) (x).size())
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define bp(x) __builtin_popcount(x)
#define INF numeric_limits<long long int>::max()
#define frp freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define forit(it, s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it++)

const int maxn = (int)1e6;
const int mod = (int)1e9 + 7;

using namespace std;

int n;
int a[1010];

main(){
    scanf("%d",&n);
    for(int i=1; i <= n; i++){
        int x; scanf("%d",&x);
        a[x]++;
    }
    for(int i=0; i <= 10; i++){
        if(a[i] > 0)
            printf("%d %d ",a[i], i);
    }


    return 0;
}

Firstly my solution got TLE#4. Next, I make mew solution with new idea, and got WA#4.
Also, my 2nd solution makes mistake in this test:

4
0 -2 -1 0
9 2 -9 -2
-4 1 -1 1
-1 8 10 -2

answer is 18.

It can help someone.
P.S. forry for bad English.

What's wrong? Here is my code:

var a,b,c:array [1..3,1..3] of string; d,s,i,j,j1,n,p:longint; x:string;
begin
a[1,1]:='Alice';
a[1,2]:='Ariel';
a[1,3]:='Aurora';
a[2,1]:='Phil';
a[2,2]:='Peter';
a[2,3]:='Olaf';
a[3,1]:='Phoebus';
a[3,2]:='Ralph';
a[3,3]:='Robin';
b[1,1]:='Bambi';
b[1,2]:='Belle';
b[1,3]:='Bolt';
b[2,1]:='Mulan';
b[2,2]:='Mowgli';
b[2,3]:='Mickey';
b[3,1]:='Silver';
b[3,2]:='Simba';
b[3,3]:='Stitch';
c[1,1]:='Dumbo';
c[1,2]:='Genie';
c[1,3]:='Jiminy';
c[2,1]:='Kuzko';
c[2,2]:='Kida';
c[2,3]:='Kenai';
c[3,1]:='Tarzan';
c[3,2]:='Tiana';
c[3,3]:='Winnic';
readln (n);
s:=0;
p:=1;
for i:=1 to n do begin
readln (x);
for j:=1 to 3 do
for j1:=1 to 3 do begin
If x=a[j,j1] then d:=1;
If x=b[j,j1] then d:=2;
If x=c[j,j1] then d:=3;
end;
s:=s+abs (p-d);
p:=d;
end;
writeln (s);
end.

#include <iostream>

using namespace std;

int main()
{
int** m;
long n;
cin>>n;
m = new int*[n];
for (long i=0; i<n; i++)
{
m[i]=new int[2];
m[i][1]=-1;
}

for (long j=0; j<n; j++)
{
    int g, t;
    cin>>g>>t;
    if (j==0)
    {
        m[0][0]=g;
        m[0][1]=t;
        continue;
    }
    else
    for (long i=0; i<=j; i++)
    {

        if (m[i][1]==t)
        while (m[i][1]==t && i<j)
        i++;
        else
        if (m[i][1]>t && m[i][i]!=-1) continue;

            if (i==j)
        {
            m[j][0]=g;
            m[j][1]=t;
            break;
        }

{
            for (long p=j; p>i; p--)
            {
            m[p][0]=m[p-1][0];
            m[p][1]=m[p-1][1];
            }
            m[i][0]=g;
            m[i][1]=t;
            break;
            }
            break;
        }
}
for (int i=0; i<n ;i++)
{
cout<<m[i][0]<<" "<<m[i][1]<<'\n';
}
delete[] *m;
return 0;
}

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
    char s[1000];
    long l=0;
    cin.getline(s, sizeof(s));
    for(int i=1; i<=strlen(s); i++)
    switch (s[i])
    {
        case 'a':l++; break;
        case 'd':l++; break;
        case 'g':l++; break;
        case 'j':l++; break;
        case 'm':l++; break;
        case 'p':l++; break;
        case 's':l++; break;
        case 'v':l++; break;
        case 'y':l++; break;
        case '.':l++; break;
        case 'b':l+=2; break;
        case 'e':l+=2; break;
        case 'h':l+=2; break;
        case 'k':l+=2; break;
        case 'n':l+=2; break;
        case 'q':l+=2; break;
        case 't':l+=2; break;
        case 'w':l+=2; break;
        case 'z':l+=2; break;
        case ',':l+=2; break;
        case ' ':l++; break;
        case 'c':l+=3; break;
        case 'f':l+=3; break;
        case 'i':l+=3; break;
        case 'l':l+=3; break;
        case 'o':l+=3; break;
        case 'r':l+=3; break;
        case 'u':l+=3; break;
        case 'x':l+=3; break;
        case '!':l+=3; break;
        default: l++;
    }
    cout << l;
}

#include <iostream>
using namespace std;
void main()
{
    int a, b;

    cin >> a >> b;

    cout << "The sum is: " << output << endl;
}

Edited by author 31.10.2015 10:48

I have WA4. I don't understand why? Give me some tests please, which will help me understand my mistake.
Thank you.

Perhaps #4 is about a test without the back wall. Here come some questions:
1. Should we assume the back wall to be located in the n+1 - th ?
2. It seems that the second scanning is from the back. So the FIRST maximum refers to the one who is leftmost or the rightmost?