1. This problem can be reduced to the following:
      Given X, find two vertices such that path weight is equal to X.
2. It probably cannot be solved with naiive DP. Use centroid decomposition.
3. std::unordered_map gets TL while std::map gets AC.

i have some problems with (), please give answers for some other cases

What it is it? My program overcome all my tests,but system(TIMUS) written me about WA#9!
 WHY?

Don't be afraid of geometry, this problem is really easy

Can you help me?

#include<iostream>

using namespace std;
int sum(int temp);
int main(){
    int a[6];
    scanf("%1d%1d%1d%1d%1d%1d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5]);
    int first=a[0]*100+a[1]*10+a[2];
    int second=a[3]*100+a[4]*10+a[5];
    if(a[3]==9&&a[4]==9&&a[5]==9){
        cout<<"No";
        return 0;
    }
    int pre=second-1;
    int next=second+1;
    if((sum(pre)==(a[0]+a[1]+a[2]))||(sum(next)==(a[0]+a[1]+a[2]))){
        cout<<"Yes";
    }else{
        cout<<"No";
    }

return 0;
}

int sum(int temp){
    int count=0;
    do{
        count+=(temp%10);
        temp/=10;
    }while(temp/10!=0);
    count+=temp;
    return count;
}

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

class team{
    public:
        string id;
        int n;
};

bool comp(const team &t1,const team &t2);

int main(){
    int k;
    cin>>k;
    vector<team> T;
    while(k--){
        string temp;
        int num;
        cin>>temp>>num;
        team t;
        t.id=temp;
        t.n=num;
        T.push_back(t);
    }
    stable_sort(T.begin(),T.end(),comp);
    for(int i=0;i<T.size();++i){
        cout<<T[i].id<<" "<<T[i].n<<endl;
    }

return 0;
}

bool comp(const team &t1,const team &t2){
    return t1.n>t2.n;
}

Why Wring answer?
import java.util.Scanner;

public class T1654 {
    public static char[] arr;
    public static String out = "";
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        String inp = sc.next();
        sc.close();
        arr = new char[inp.length()];
        arr = inp.toCharArray();
        format();
        if(out != "") {
            System.out.println(out);
        } else {
            System.out.println("");
        }

    }

    public static void format() {
        boolean isFormat = false;
        int l = arr.length;
        for(int i = 0; i< l-1; i++) {
            if(arr[i] == arr[i+1]) {
                arr[i] = 0;
                arr[i+1] = 0;
                isFormat = true;
                } else if(arr[i] != arr[i+1] & arr[i] != 0){
                out += arr[i];
            }
        }
        out+= arr[arr.length-1];
        if(isFormat == true) {
            arr = new char[out.length()];
            arr = out.toCharArray();
            out = "";
            format();
        }
    }

}

получил АС, долго искал где долбанный баг, т.к. ТЕСТ 22 выдавал ошибку по времени, если у вас такая же проблема и вы используете преобразование double в int то в 22 тесте точность P и Q 3 знака, поэтому ТЛ и выдает, вероятно на вход подаются какие нибудь числа 22.333 и 22.334
удачи

code like:

size_t out = 0;
printf ("%zu\n", out);

result - > wa#1 (((

if i use

printf ("%d\n", out);

all  test accepted (( Why?
https://wandbox.org/ & gcc 7.1  - The first version works correctly (((

i have ac in 0.187 s, but on this test my program runs at least 0.9:

35 123
2 3 4 5 6 0 8 9 10 11 0 13 14 15 0 17 18 0 20 21 0 23 0 25 0 27 0 0 0 0 0 0 0 0 0

I've got accepted after adding the condition of end of file in creating new line. May be, it will be useful.

Edited by author 31.01.2016 15:50

Edited by author 31.01.2016 15:50

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int main(){
    int k;
    cin>>k;
    vector<int> line1;
    vector<int> line2;
    vector<int> sum;
    while(k--){
        int m,n;
        cin>>m>>n;
        line1.push_back(m);
        line2.push_back(n);
    }
    bool plus_1=false;
    for(int i=line1.size()-1;i>=0;i--){
        int temp=line1[i]+line2[i];
        if(plus_1){
            temp++;
        }
        if(temp>=10){
            plus_1=true;
            temp-=10;
        }else{
            plus_1=false;
        }
        sum.push_back(temp);

    }
    for(vector<int>::reverse_iterator iter=sum.rbegin();iter!=sum.rend();++iter){
        cout<<*iter;
    }
return 0;
}

I've been wa on this case for a long time:
abaabaaba
The right answer is abaabaabaaba

Can anybody please give me cases for Test2 ?

sandro
answer is 5

Really fast.

for (int k = 0; k < n; ++k) {
  for (int i = graph[k]._Find_first(); i < n; i = graph[k]._Find_next(i)) {
    graph[k] |= graph[i];
  }
}

I created array p[1..10000000] of boolean; p[i]=true when exist i in input. But i can t prove that if length of input <=100000 then absent numb <=10000000

Edited by author 11.02.2019 20:58