Не слушайте тех кто пишет что убрал проверку за границы и зашла и все в таком духе.
Возможно у вас такая же ошибка как и у меня, а то есть когда ввели новые координаты я проверял только то что может ли эту новую координату кто то ударить, а надо было еще проверить что может ли новая координа ударить кого то из врагов

I use some shamanism ang get AC. Lets see to the next test:

6
0 20
10 50
30 100
60 120
70 130
110 140

my AC prog outputs:

2
10 50
110 140

but right answer of course is:

3
0 20
30 100
110 140

Your tests must be stronger

Извините, что по-русски, надеюсь, администраторы знают этот язык.

В условии сказано, что длина последовательности задается в первой строке каждого теста.
В примерах на форуме длина общая для всех, в каждом тесте задается только количество вопросов.
Моя AC-программа действует по форуму: читает длину один раз, а потом только количество вопросов для каждого теста.

Хорошо бы привести текст задачи в соответствие с реальностью.
Совсем хорошо было бы дать в примере два теста, чтобы сделать структуру входных данных действительно ясной.

def word_to_num(word):
    d=
{'a':'2','b':'2','c':'2','d':'3','e':'3','f':'3','g':'4','h':'4','i':'1','j':'1','k':'5','l':'5','m':'6','n':'6','o':'0','p':'7','q':'0','r':'7','s':'7','t':'8','u':'8','v':'8','w':'9','x':'9','y':'9','z':'0'}
    s=''
    for i in word:
        s+=d[i]
    return s
from itertools import permutations
while True:
    num_dict={}
    num_list=[]
    word_list=[]
    num = input()
    if num=='-1':
        break
    loop_len = int(input())
    for i in range(loop_len):
        word_list.append(input())
    for item in word_list:
        num_dict[word_to_num(item)]=item
        num_list.append(word_to_num(item))
    perm = permutations(num_list,2)
    x=0
    for k,v in list(perm):
        if k+v==num:
            print(num_dict[k]+' '+num_dict[v])
        else:
            x += 1

    if x==loop_len*(loop_len-1):
        print('No solution')

For Time limit exceeded:
If you get time limit exceeded, it must mean that you're trying to modify the
input string (i.e. trying to delete i-th chars). Try creating a new OUTPUT string
without modifying the INPUT string.

Note: if you're using a high language such as java or C#, try using the StringBuilder
object to build your output string.

For Wrong Answer(WA):
If you get WA, try some of these test cases:

Input: "dddd"
Output: ""

Input: "ddddd"
Output: "d"

Input: "abba"
Output: ""

Input: "abbbbac"
Output: "c"

Input: "wliisddsmmleeddw"
Output: ""

Input: "avcbbffcv"
Output: "a"

Hope this helps.

Edited by author 12.07.2012 00:47

Hi, I writed this post because I wanted to share with you some of the test cases* that helped me a lot at the time. I hope it helps you :)

--> Test 1:

1
A B C

--> Answer:

A undefined
B undefined
C undefined

------------------------------

--> Test 2:

5
Isenbaev A B
A B C
D Q P
C H N
G N P

--> Answer:

A 1
B 1
C 2
D 5
G 4
H 3
Isenbaev 0
N 3
P 4
Q 5

------------------------------

--> Test 3:

13
Fominykh Isenbaev BBB
BBB CCC AAA
Ayzenshteyn Oparin Samsonov
Ayzenshteyn Chevdar Samsonov
Dublennykh Fominykh Ivankov
Burmistrov Dublennykh Kurpilyanskiy
Cormen Leiserson Rivest
Oparin AA AAA
Isenbaev Oparin Toropov
AA DD PP
PP QQ RR
RR SS TT
TT Toropov Oparin

--> Answer:

AA 2
AAA 2
Ayzenshteyn 2
BBB 1
Burmistrov 3
CCC 2
Chevdar 3
Cormen undefined
DD 3
Dublennykh 2
Fominykh 1
Isenbaev 0
Ivankov 2
Kurpilyanskiy 3
Leiserson undefined
Oparin 1
PP 3
QQ 4
RR 3
Rivest undefined
SS 3
Samsonov 2
TT 2
Toropov 1

------------------------------

--> Test 4:

3
Isenbaev A B
A B C
A Q W

--> Answer:

A 1
B 1
C 2
Isenbaev 0
Q 2
W 2

------------------------------

*(All test cases were made by other users, so you'll probably found them in other Posts/Topics/Discussions).

Acknowledgments to:
-> "Vit Demidenko" & "Nathalie"
-> "Pavel Nikolov"
-> "mccolt89"
-> "Megakrit"


-How I solved it:

I solved this problem in Java, for that, I used:

->1 java.util.Scanner
->1 java.util.HashMap
->1 java.util.TreeMap
->3 java.util.HashSet
->1 java.util.LinkedList
->? java.util.StringTokenizer (or you can just simply use the Scanner method "next()").

(The structures can be declared inside loops or other structures).

Hope all this helped you :)


Edited by author 08.04.2020 00:48

Edited by author 08.04.2020 00:49

Test in you code example like this
4 3
1 5 6
Wrong Answer is 0.
Thrue Answer is 3.
Be confident!!!

#include<iostream>
using namespace std;
int main(){
    int k, n, arr[101],x=0;
    cin>>k>>n;

    for(int i=0; i<n; i++){
        cin>> arr[i];

    }
    for(int i=0;i<n;i++){

        if(k>=arr[i]){
            if(i != n-1){
                x = x + arr[i+1];
                arr[i+1] = x;

                x = 0;

            }
            else{
                x = 0;
            }


        }
        else{
            if(i != n-1){
                x = (arr[i]-k) + arr[i+1];
                arr[i+1] = x;


            }
            else{
                x = arr[i] - k;
            }
        }
    //    printf("X : %d\n",x);
    }
    if(x<=0)
    cout<<0;
    else
    cout<<x;

}

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector <vector <int>> a(n, vector <int> (n));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j];
        }
    }
    int ans = a[0][0];
    for (int leng = 1; leng <= n; leng++){
        for (int up = 0; up <= n-leng; up++){
            vector <long long> sumv(n);
            for (int j = 0; j < n; j++) {
                for (int i = up; i < up + leng; i++) {
                    sumv[j] += a[i][j];
                }
            }
            int sum = 0;
            for (int r = 0; r < n; ++r) {
                sum += sumv[r];
                ans = max(ans, sum);
                sum = max(sum, 0);
            }
        }
    }
    cout << ans;
}

has anybody got WA8 too? I tried all tests in forum but it didn't help :(

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main() {
    string c = "a";
    string word;
    for (;;) {
        c = getc(stdin);
        if (c[0] == EOF) {
            reverse(word.begin(), word.end());
            cout << word;
            break;
        }
        if (((c[0] >= 'a') && (c[0] <= 'z')) || ((c[0] >= 'A') && (c[0] <= 'Z'))) {
                word.append(c);
        }
        else {
            reverse(word.begin(), word.end());
            cout << word << c;
            word.clear();
        }
    }
}

Что_такой_тест_13?

instead map use unordered_map

once I've solved this exercise, but know a vontto make my algorithm better. I've forced whith WA#12. Could you tell me this test?

#include <iostream>
#include <vector>
#include <iterator>
using namespace std;
int n = 163841;
vector<char> prime(n + 1, true);
void eratosphen() {
    prime[0] = false;
    prime[1] = false;
    for (int i1 = 2; i1 <= n; ++i1)
        if (prime[i1])
            if (i1 * 1ll * i1 <= n)
                for (int j1 = i1 * i1; j1 <= n; j1 += i1)
                    prime[j1] = false;
}
int main() {
    eratosphen();
    vector <long long> list(15001); int i = 2;
    for (int il = 1; il <= 15000;) {
        for (int ip = 2; ip < prime.size(); ip++) {
            if (prime[ip]) {
                list[il] = ip;
                il++;
            }
        }
    }
    int k;
    cin >> k;
    for (int i0 = 0; i0 < k; i0++){
        int m;
        cin >> m;
        cout << list[m] << '\n'; \
    }
}

SORT!!!

Sort the quarters that can be crossed.And then dp.

The state transition equation is
F[i]=Max{ F[j] + 1 | P[j].x<P[i].x and P[j].y<P[i].y }

The maximum of F[i] is the number of the the quarters that should be crossed.

Then you can work out the answer.

Edited by moderator 18.08.2020 02:22

We could solve it by Theorem on the sum of four squares, I wonder how to solve it by DP? If you know, please help me, thank you!

I failed on 13 test and I can't found error. Please give a 13 test.