3 6 8 20 30 40
7
1 6
3
7
8
13
15
23
//answer: 80

3 6 8 20 30 40
7
3 6
3
7
8
13
15
23
//answer: 40

3 6 8 20 30 40
7
2 5
3
7
8
13
15
23
//answer: 60

1 2 3 1 1000 1000000000
2
1 2
1
//answer: 1

1 2 3 1 1000 1000000000
2
2 1
1
//answer: 1

#include<iostream>
#include<conio.h>
#include<string>
using namespace std;
char unCAPS(int state, char symbel)
{
    char out_symb;
    if (((int)symbel >= 65 && (int)symbel <= 90) && state ==0)
    {
        out_symb = (char)(symbel + 32);
        return out_symb;
    }

    if (((int)symbel >= 97 && (int)symbel <=122 ) && state == 1)
    {
        out_symb = (char)(symbel - 32);
        return out_symb;
    }

    return symbel;
}

int main()
{
    char str[10000] = { NULL }, str2[10000] = { NULL };
    int state = 1;
    cin.get(str, 10000);
    for (size_t i = 0; i < strlen(str); i++)
    {
        str2[i] = unCAPS(state, str[i]);
        if ((int)str[i] != 32)
        {
            state = 0;
        }
        if ((int)str[i] == 46 || (int)str[i] == 33 || (int)str[i] == 63)
        {
            state = 1;
        }

    }
    for (size_t i = 0; i < strlen(str2); i++)
    {
        cout << str2[i];
    }

    return 0;
}

Try this test:
3 1
1000 0
1000 6
100 10
2 5

answer is:
1985.0

#include <iostream>
#include <cmath>
using namespace std;
void main(){
    int a[100001];
    char s[100001];
    gets_s(s);
    int max = 0;
    for (int i = 0; i <= strlen(s); i++)
    {
        if (s[i] >= 'A'&& s[i] <= 'Z'){
            a[i] = s[i] - 55;
            if (max < a[i]){
                max = a[i];
            }
        }
        else{////if digits
            a[i] = s[i] - 48;
            if (max < a[i]){
                max = a[i];
            }
        };
    };
    int k;
    long long res;
    max++;
    bool found = 1;
    for (k = max; k <= 36; k++)
    {
        res = 0;
        for (int i = strlen(s) - 1; i >= 0; i--)
        {
            res = res + pow(k, i)*a[i];
        }
        if (res % (k - 1) == 0){
            found = 0;
            break;
        }
    }
    if (found==0){
        cout << k;
    }
    else{
        cout << "No solution.";
    }
    system("pause");
}
I think that problem in large numbers in tests, am I right?
Edited by author 09.07.2016 22:05

Edited by author 09.07.2016 22:26

Edited by author 21.05.2017 17:19

See on one iteration of "for":
let be x and y on start:x1 and y1.
then after each change it'll be for x1 x2...x3; for y1 y2...y3;
so you have to find out x3 and y3 through x and y on start(x1 and y1);
Something like that x3=y1.
Good luck and sorry if my English is not enough good for explaining it)

5
777 313 465 99 1

in the test above, how Daenerys  can win ??? While we can divide:
#Daenerys  777 -> 3 heaps (259)
- 313 -> can't divide
#Stannis  465 -> 3 heaps (155)
#Daenerys 99  -> 3 heaps (33)
#Stannis  33 -> 3 heaps (11)
- 1   -> can't devide
==> Stannis  is the winner

I don't understand how the answer can be 15.

Since the first 2 people got 0 correct predictions and the last person got 1 correct prediction, shouldn't the amount paid only be 5?

dp[i][j] - represents number of staircases with i cubes and with the height of last block j.
Answer is dp[n][1] + dp[n][2] + dp[n][3] ... dp[n][n].

At first check all tests with n<=4. If all of them cheked already then try to check all cells for n=8 (1/1; 1/2; 1/3; 1/4; 2/2; 2/3....)and for n=9. Don't be afraid you'll find your mistake on first tests))

Edited by author 10.07.2016 14:33

can someone give me some tips or tests?

#include <iostream>
#include <math.h>
#include <iomanip>

using namespace std;

int main()
{
    double a[100000];
    int i, n;
    cin >>n;
    for (i=1; i<=n; ++i)
    {
        cin>>a[i];
    }
    for (i=1; i<=n; ++i)
    {
        a[i] = sqrt(a[i]);
    }
    for (i=n; i>=1; --i)
    {
        cout << fixed << setprecision(4) << (a[i]) << endl;
    }
}

What is the criterion of optimality of the partition ?

По условию может быть так(.-пробел):.........aaa.bbb.ссс.
                                    .....xxx.aaa.bbb.ccc.
                                    или
                                    .........aaa.bbb.ссс........
                                    .....xxx.aaa.bbb.ccc........???????
И вообще,как в ЭТОЙ задаче считывать данные(на Паскале)?

Edited by author 08.07.2016 21:27

I tested my prog but I not found mistake

public class AB {
public static void main(String [] args) {
int a = 1;
int b = 5;
System.out.println((a+b));
}
}