I am getting a wrong answer in it. Can anyone please help me?

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int m,n,a[50000],b[50000],i,j,p,c;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    scanf("%d",&m);
    for(i=0;i<m;i++)
    {
        scanf("%d",&b[i]);
    }
    for(j=0;j<n;j++)
    {
        for(p=0;p<m;p++)
        {
            c=a[j]+b[p];

            if(c==10000)
            {
                 printf("YES");
                 exit(0);
            }
        }
    }
    printf("NO");
    return 0;
}

import java.util.Scanner;

public class Factorials {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String k = in.next();
        char a[] = k.toCharArray();
        int length = a.length;
        function(n, length);
        in.close();
    }

    private static void function(int n, int length) {
        int t = n % length;
        int c = (t == 0 ? length : t);
        int sum = n;
        for (int i = 1;; i++) {
            sum = sum * (n - i * length);
            if ((n - i * length) == c) {
                System.out.println(sum);
                break;
            }
        }
    }
}

Any ideas?

http://ideone.com/ZFtToc

I don't see what can be wrong with it...

In seems that in C++ for set-object no operator for
standard container inquiry : N-tn element,  with log N complexity,  but should be.
With such operator problem 1439 would be very simple.
In reality  was necessary to program my own set class based on red-black trees. And N-th element to realize was very simple(hard work was balancing).
It,s not well that tree-structures of standard containers are hidden  for uses in C++.

My Code:
#include <bits/stdc++.h>
using namespace std;

int main()
{
    //freopen("in.txt","r", stdin);

    unsigned long long int n,k,x,j; cin >> n;
    for(int i=1; i<=n; i++)
    {
        cin >> k;
        x=1;
        j=0;
        while(j<k)
        {
            j+=x;
            if(j<k) x++;
        }
        x--;

        if(j-k < x) cout << 0 << " ";
        else cout << 1 << " ";
    }
    cout << endl;

    return 0;
}

Edited by author 26.04.2016 00:41

My program passed all tests, which I found here or invented myself, but I have WA4.
Can anybody help me find a mistake or give some special tests? I'd be very grateful...

--
P.S. Already solved. This test helped me:
5
RGBRB

Edited by author 27.04.2016 20:49

Can anyone tell me  the "main idea" of test 4 or type test. I got AV. Need some reason where is my mistake (to my mind - arrays, but need help). Thanks.

1) First of all you have to calculate maximum of tickets the teacher can buy ((amount of money) div (price of one ticket))
2) Next step you have to find minimum of tickets. You must pick up one banknote with minimum denomination and calculate amount again: in sample test you get 0 1 0 0 0 0 sequence. Then calculate amonut of money again. So, now you can get minimun of tickets ((new amount of money) div (price of one ticket) + 1).
3) So, now you have minimum and maximum of tickets. Just output the number of possible answers in the first line (tMax - tMin + 1) and the variants in ascending order (for i:=tMin to tMax do write(i,' ');

Good luck and get AC!

When I run the code it gives me a runtime error. Why? Python3 for problem 1011.

from math import ceil, floor

a = float(input())
b = float(input())

if a>b:
    a, b = b, a

num = 1
while ceil((a/100)*num) != floor((b/100)*num):
    num += 1

print(num)

Edited by author 06.05.2016 11:07

This is my code!!!
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
 vector <string> str;
 vector <string> str1;
int main()
{
    char k;
 string s;
 vector <string> str;
 vector <string> str1;
 int n;
 cin>>n;
 for(int i=0; i<n; i++)
 {
     cin>>s;
     str.push_back(s);
 }
 cin>>k;

 for(int i=0; i<n; i++)
 {
     if(str[i][0]==k) str1.push_back(str[i]);
 }

 sort(str1.begin(),str1.end());
 for(int i=0; i<str.size(); i++)
 {
     cout<<str1[i]<<endl;
 }
 return 0;
}

Can somebody give me some test with corect correct output, cause it is rather hard to calculate answer on paper.

AC now, however good test will be good for other solvers.

Edited by author 28.10.2007 01:48

I didn't know anything about combinations but now I know, so I can give link(explaining on Russian) http://www.mathprofi.ru/zadachi_po_kombinatorike_primery_reshenij.html
Sorry if I have some mistakes with English :)

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;

int shkaf(string name) {
    string mas1[9] = { "Alice", "Ariel", "Aurora", "Phil", "Peter", "Olaf", "Phoebus", "Ralph", "Robin" };
    string mas2[9] = {"Bambi", "Belle", "Bolt","Mulan", "Mowgle", "Mickey", "Silver", "Simba", "Stitch" };
    string mas3[9] = { "Dumbo", "Genie", "Jiminy", "Kuzko", "Kida", "Kenai", "Tarzan", "Tiana","Winnie" };
    for (int i = 0; i < 9; i++) {
        if (name == mas1[i])return 1;
        if (name == mas2[i])return 2;
        if (name == mas3[i]) return 3;
    }
}

int main() {
    int N, s=0,a1=1,a2=1;
    string name;
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin>>name;
        a2=shkaf(name);
        s += abs(a2 - a1);
        a1 = a2;
    }
    cout << s;

    return 0;
}

/*
Le Van Manh
Hanoi University Of Science And Technology
Viet Nam
*/
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
int readAddressToValue(){
    int v = 0;
    string a;
    cin>>a;
    if(
        a=="Alice"||a=="Ariel"||a=="Aurora"||
        a=="Phil"||a=="Peter"||a=="Olaf"||
        a=="Phoebus"||a=="Ralph"||a=="Robin"){
        v = 0;
    }else
    if(
        a=="Bambi"||a=="Belle"||a=="Bolt"||
        a=="Mulan"||a=="Mowgli"||a=="Mickey"||
        a=="Silver"||a=="Simba"||a=="Stitch"){
        v = 1;
    }else{
        v = 2;
    }
    return v;
}
int main(){

    int address0;
    int address1;
    int steps = 0;
    int n;
    cin>>n;

    address0 = readAddressToValue();
    steps+= address0 - 0;
    for(int i=1;i<n;i++){
        address1 = readAddressToValue();
        steps += abs(address1 - address0);
        address0 = address1;
    }

    cout<<steps;
    return 0;
}

Edited by author 20.02.2016 13:17

You just need an array with 100 elements so in every position M, you add the ID using a dynamic structure such as "vector" in C++. (Theoretical complexity O(n)) Now you go from 100 to 0 and then print every element from each position M.

Edited by author 23.04.2016 15:06

if you have a better algo please tell to me.

plz help!!!!!!!!!!!!!!!

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
    int N,M,a,b;
    vector <int> t;
    vector <int> s;
    vector <int> st;
    cin>>N;
    for(int i=0; i<N; i++)
    {
        cin>>a;
        t.push_back(a);
    }
    cin>>M;
    for(int i=0; i<M; i++)
    {
        cin>>b;
        s.push_back(b);
    }
    for(int i=0; i<M; i++)
    {
        for(int j=0; j<N; j++)
        {
            if(s[i]==t[j])  st.push_back(s[i]);
        }
    }
    sort(st.begin(),st.end());
    cout<<st.size();
    return 0;
}

import java.util.Scanner;

public class SMSspam2 {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        String s =in.nextLine();
        char a[] = s.toCharArray();

        int count = 0;
        for(int i=0;i<s.length();i++){
            count = count+getCount(a[i]);
        }
        System.out.println(count);
        in.close();
    }

    private static int  getCount(char a) {
        int count =0;
        if (a == 'a' || a == 'd' || a == 'g' || a== 'j' || a== 'm' || a == 'p' || a == 's' || a== 'v' || a == 'y'
                || a== '.' || a== ' ') {
            count++;
        }
         if (a== 'b' || a == 'e' || a== 'h' || a == 'k' || a == 'n' || a == 'q' || a== 't' || a== 'w' || a== 'z'
                || a == ',') {
            count = count + 2;
        }
        if (a== 'c' || a== 'f' || a == 'i' || a == 'l' || a== 'o' || a== 'r' || a == 'u' || a== 'x' || a == '!') {
            count = count+ 3;
        }
        return count;
    }
}