Seems an easy problem, but I can't understand it...

test it your self

after deleting this code, please
explain me why WA twoalias[animal]inbox[youknow]ru

#include <stdio.h>
#include <memory.h>


struct elt
{
    int c;
    short i;
};

elt T[100001];

int main()
{
    memset(T,0,sizeof(T));
    int N;
    scanf("%i",&N);
    short i;
    for (i=1;i<=N;i++)
    {
        int S,C;
        scanf("%i %i",&S,&C);
        if (C>T[S].c)
        {
            T[S].c=C;
            T[S].i=i;
        }
    }
    int j,ch=0;
    for (j=0;j<100001;j++) if (T[j].c>0) ch++;
    printf("%i\n",ch);
    for (j=0;j<100001;j++) if (T[j].c>0) printf("%i ",T[j].i);
    return 0;
}

Edited by author 29.08.2006 20:03

Just use double type instead int

it is possible that the collection be empty after some operations!
gcd(empty set)=1.

Edited by author 31.03.2012 15:04

1. If time limit exceeded, use a better sorting algorithm. In my case, insertion sort spent barely over a second, while merge sort spent only 0.6 of a second.

2. used long long. Also, before multiplying a long long and an int together, convert the int first.

Check the way how you find the upper radius if h>d.

#include <iostream>
#include <math.h>
#include <vector>
#include <stdlib.h>

using namespace std;
int main() {
    vector<double> v;
    int v1;
    while((cin>>v1))
    {
        v.push_back(v1);
    }
    int size = v.size();
    int first = v[0];
    int last = v[size-1];
    for(int i = size-1; i>=0; i--)
    {
        double a = sqrt(v[i]);
        printf("%.4lf\n", a);
    }
    return 0;
}

All forum tests passed, but wa:( Million thanks to someone who will help!

Can someone tell me how I can solve thi problem using ordered set?

What is test №36? I don't understand why my programm crash on test.

Т.е. у этой задачи предполагается написание перебора с отсечениями? Последовательно добавляем символы в конец строки и если не смогли найти палиндрома, который бы являлся суффиксом новой строки, то дальше не перебираем?

Hint:
We must find pair for the first character or create it.
If S[0] is ']' or ')', then we must add '[' or '(' in begin.
If S[0] is '[' or '(', then we can find pair in S or add ']' or ')' after S[0].

Edited by author 26.06.2020 19:51

pleeeeesseee give test

9 3
1 5
1 4
2 7
2 5
3 6
5 9
6 9
7 8
1 4
2 4
9 4

Answer:
8 (or 3)
3
8

Hello!

I'm successfully passing first cases and also cases from there https://acm.timus.ru/forum/thread.aspx?id=41894&upd=636800027260191369 , but what about test #4. could you please share some sort of example for that?

Thanks is advance!

3
5 5
5 1
5 7

ans
1
2 3

#include <bits/stdc++.h>

#define gif(a,b) ((a)/(b)+((a)%(b)?1:0))
#define float long double
#define int long long
//#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define w(x) cout<<(#x)<<" is "<<(x)<<"\n";
using namespace std;

int32_t main()
{
    //IOS
    float a, r;
    cin >> a >> r;
    float theta = acos(a/(2*r));


    float s = sin(theta);

    float pi = acos(-1);
    if(a/2 > r)
        printf("%0.3Lf",pi*r*r);
    else if(r > a/pow(2, 0.5))
        cout << a*a;
    else
    cout << (pi*r*r) - 4*(r*r*theta - a*r*s/2) ;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int primes[300001],nprime;
int mark[1000002];
#define MAX_SIZE 1000005

void sieve(){
int i,j,limit = sqrt(MAX_SIZE*1.0)+2;
mark[1]=1;
for(i=4;i<=MAX_SIZE;i+=2)mark[i]=1;
primes[nprime++]=2;
for(i=3;i<=MAX_SIZE;i+=2){
    if(!mark[i]){
        primes[nprime++]=i;
        if(i<=limit){
            for(j=i*i;j<=MAX_SIZE;j+=i*2){
                mark[j]=1;
            }
        }
    }
}
}

int main(){
sieve();
int n;
cin>>n;
while(n--){
    int x;
    cin>>x;
    cout<<primes[x-1]<<endl;
}

}

Edited by author 26.10.2013 22:11