I THINK THAT MY PROGRAM MUST HAVE AC BUT WA3
WHAT THE 3 TEST????
PLEASE POST TEST №3!!!!!!!

dic={'a':1,'d':1,'g':1,'j':1,'m':1,'p':1,'s':1,'v':1,'y':1,'.':1,' ':1,
      'b':2,'e':2,'h':2,'k':2,'n':2,'q':2,'t':2,'w':2,'z':2,',':2,
      'c':3,'f':3,'i':3,'l':3,'o':3,'r':3,'u':3,'x':3,'!':3}

m=input()
total=0
for i in range(len(m)):
    total+=dic[m[i]]
print(total)

a,b=input().split()
n=int(a)
k=len(b)
temp=n
t=k
while temp-k>0:
        n*=(temp-k)
        k+=t

print("%d"%n)

rt

i think test set is not full. my AC soulution uses floating arithmetic and does not work correctly on tests like
3
0 0 0
1 1 1
100000 100000 99999
right answer should be 2?

Input:
10 3
9 9 9

Output:
7

What is T17 ?

A nice task for implementation, but the description is quite tricky sometimes.

What if Program Counter went out of last instruction? - it is ok, you can think about it as normal termination by END command
Is NOP count towards overall operations count? - no, empty lines and only-label lines do not contribute to the overall number of executed commands
Are labels case sensitive? - no
Are operands case sensitive? - no
Are variable names case sensitive? - yes
What about int32 overflow? - guaranteed that there are no overflows
What about % operation? - the result of A % B operation is such C that 0 <= C < B and exist some integer Q: A = Q * B + C
What if 10_000_000th command is END command? - The end command terminates the program
Can variable name start from digit? - no

#include <iostream>
#include <math.h>
using namespace std;

void square()
{
    long long int n;
    scanf("%lld", &n);
    if(n!=-1)
        {
            square();
            printf("%.4f\n", sqrt(n));
        }

    return;

}
int main()
{
    square();

    return 0;
}

My program works with tests like this:

4

0 00 0

10 11

And my program give right answers but WA#1. That my code, whats wrong??

#include <bits/stdc++.h>

using namespace std;

int n;
string ans,q;

int main()
{
    cin >> n;

    while (!cin.eof()) {                                     //while not eof reading
        getline(cin,q);                                      //read 1 line
        ans = "";

        for (int i = 0; i < q.length(); i++)                 //remove all except 0 and 1
            if (q[i] == '0' || q[i] == '1')
                ans += q[i];


        if (ans.length() < n-1 || ans.length() > n+1)         //if its empty continue
            continue;

        int pos_sum = 0;

        for (int i = 0; i < ans.length(); i++) {              //count sum of positions
            if (ans[i] == '1') {
                pos_sum += i+1;
            }
        }

        if (ans.length() > n) {                              //if there 1 unnecessary element check all
            bool fg = false;                                 //elements and remove him

            for (int i = 0; i < ans.length(); i++) {
                int loc_pos_sum = 0;

                if (fg) {
                    cout << ans[i];
                    continue;
                }

                for (int j = 0; j < ans.length(); j++) {
                    if (j == i)
                        continue;

                    if (ans[j] == '1')
                        loc_pos_sum += j+1 - int(j >= i);
                }

                if (loc_pos_sum % (n+1) == 0) {
                    fg = true;
                } else {
                    cout << ans[i];
                }
            }
        } else if (ans.length() < n) {                                 //if there 1 removed element
            bool fg = false;                                           //check all positions for him

            for (int i = 0; i <= ans.length(); i++) {
                int loc_pos_sum = 0;

                if (fg) {
                    cout << ans[i];
                    continue;
                }

                for (int j = 0; j < ans.length(); j++) {
                    if (ans[j] == '1') {
                        loc_pos_sum += j+1 + int(j >= i);
                    }
                }

                if (loc_pos_sum % (n+1) == 0) {
                    cout << 0;
                    fg = true;
                }

                if ((i+1+loc_pos_sum) % (n+1) == 0) {
                    cout << 1;
                    fg = true;
                }

                cout << ans[i];
            }
        } else {                                                          //if '0' replaced to '1' check all
            bool fg = false;                                              //elements and replace
            for (int i = 0; i < ans.length(); i++) {
                if (ans[i] == '1' && (pos_sum-i-1)%(n+1) == 0 && !fg) {
                    cout << 0;
                    fg = true;
                } else {
                    cout << ans[i];
                }
            }
        }

        cout << "\n";
    }
    return 0;
}

Edited by author 30.10.2018 16:07

For some inputs multiple answers are possible, e.g.:

4
1011

has two possible answers: 1111 and 1001. However only 1001 seems to be accepted.

Either the wording of the problem should be changed or the alternative answers should be accepted.

Somebody can give me test case 4?
I don't understand why it's wrong answer...

Edited by author 30.10.2018 11:59

#include <iostream>
#include <math.h>
using namespace std;

int Prime[15000], nPrime;
int mark[15000];

void sieve(int n)
{
    int i, j, limit=sqrt(15000)+2;

    mark[1]=1;  ///mark is not prime...so...

    for(i=4; i<=n; i+=2)
        mark[i]=1;

    Prime[nPrime++]=2;

    for(i=3; i<=n; i+=2)
        if(!mark[i])
        {
            Prime[nPrime++]=i;
            if(i<=limit)
            {
                for(j=i*i; j<=n; j+=i*2)
                    mark[j]=1;
            }
        }
}

int main()
{
    sieve(15000);
    int n;
    cin >> n;
    int arr[2000];
    //cout << Prime[n-1] << endl;

    for(int i=0; i<n; i++)
        {
            cin >> arr[i];
        }
    for(int i=0; i<n; i++)
        cout << Prime[arr[i]-1] << endl;

        return 0;
}

HHHHHH.KJDFDKJ(newline)
ADFFG

right answer

Hhhhhh.Kjdfdkj
adffg

I WA here for five times

Edited by author 25.07.2008 08:43

Edited by author 25.07.2008 08:54

subj

if you use dp, l, r can be (>30000).

Edited by author 26.10.2018 11:42

Edited by author 26.10.2018 11:42

You may just want to continue the sequence from sample, for me it was enough to get it for n = 5
{ 1, 3, 2, 6, 8, 4, 11, 5 } to start noticing the pattern.

Alternatively, just plug it into OEIS and come across A019444 with an explanation how to compute the answer :D

suppose the slope of line on the x>0 is k ,and slope of (0,0) to n
points is k1,k2,...kn then intersection point of x1==1/(k1-k),x2=1/(k2-k)...xn=1/(kn-k)
then we choose (x4-x1)/(x2-x1)==(x4'-x1')/(x2'-x1') and (x3-x2)/(x3-x4)==（x3'-x2'）/(x3'-x4')
we multiply these two equations guess what happens, yes: k is offset
then we can get (k4-k1)*(k3-k2)/((k2-k1)*(k3-k4))==(k4'-k1')*(k3'-k2')/((k2'-k1')*(k3'-k4'))
en.. this convert to string matching prolems,so suffix array can solve it

#include <iostream>
#include <cmath>

void rSqrt(void)
{
    unsigned long int n = 0;
    if (scanf("%lu", &n) != -1)
    ¦  rSqrt();
    else
    ¦   return;
    printf("%.4f\n", sqrt(n));
    return;
}

int main()
{
    rSqrt();
    return 0;
}