Could somebody who solved this problem give some tests to me? WA #7 :(

#include <stdio.h>
int main(){
    int a,b,c;
    scanf("%d",&a);
    scanf("%d",&b);
    scanf("%d",&c);
    if (a>b){
        if (b>c){
            printf("%d",c-(a*b));
        }
        else{
            printf("%d",b-(a*c));
        }
    }
    if(b>a){
        if (a<c){
            printf("%d",a-(c*b));
        }
    }
return 0;
}

I used next_permutation and get AC

Hi,

I made a countries ranklist here: http://acmtimusru.appspot.com

It lists the countries by rating, solved, users or last ac, where:

  - rating  = sum(user.rating) for user in a country

  - solved  = sum(user.solved) for user in a country

  - users   = count of users in a country

  - last ac = the most recent accepted submission of any user in a country


Also, clicking on a country gives you a ranklist of users in that specific country.

The ranklists are updated hourly.

Enjoy!

Edited by author 24.03.2020 14:51

Can anybody help

Edited by author 20.03.2020 19:35

I don't know why my solution is failing on test case 1,
I have tested against all available test cases with correct answer , but cannot pass test case 1, can anybody please tell what is test case 1.

Is it possible to change the judge id?
Thanks

What is test case 1?

import java.util.Scanner;
public class javasucc {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int h = 2, s = 10;
        for(int i=0; i<n; i++) {
            int a = in.nextInt();
            String b = in.next();
            if(b=="hungry"){
                if(a>h) {h = a;}
            }
            else if(b=="satisfied"){
                if(a<s) {s = a;}
            }
        }
        if(h >= s) System.out.println("Inconsistent");
        else System.out.println(s);
    }
}

and the problem is that program doesn't want to do if and i can't change s to a

n=int(input())
a=[]
for i in range(n):
    s=input()
    if(s[0]=="A" or s[0]=="P" or s[0]=="O" or s[0]=="R"):
        a.append(0)
    elif(s[0]=="B" or s[0]=="M" or s[0]=="S"):
        a.append(1)
    else:
        a.append(2)
t=0
for i in range(n):
    if(i+1<n):
        t=t+abs(a[i]-a[i+1])
print(t)

#include <stdio.h>
#include <map>
using namespace std;
int main() {
    int ans, sum, ost, s; long int full, i; map <long int, int> sums;
    for (i = 1; i <= 1000000000; i++) {
        sum = 0; full = i;
         do {
            ost = full % 10;
            full /= 10;
            sum += ost;
         } while (full > 0);
         sums[i] = sum;
    }
    for (s = 1; s <= 81; s++) {
        ans = 0;
        for (i = 1; i <= 1000000000; i++)
            if (sums[i] == s)
                ans++;
        printf("{%ld}, ", ans);
    }
}

it will give you {10}, {...}, ... , {...}, vector that you can use for getting AC with O(1).

Edited by author 11.03.2020 14:40

Edited by author 11.03.2020 14:44

"The numbers are arranged in non-decreasing order (0 ≤ a ≤ b ≤ c ≤ 100)."
I my opinion, reading this we assume that the input will already be sorted.

#include <iostream>
#include <string>
using namespace std;
int main() {
    int hun = 2, sat = 10, n;
    cin >> n;
    string s;
    for (int i = 0, num; i < n; i++) {
        cin >> num >> s;
        if (s == "hungry")
            if (num > hun)
                hun = num;
        if (s == "satisfied")
            if (num < sat)
                sat = num;
    }
    if (hun >= sat)
        cout << "Inconsistent";
    else
        cout << sat;
}

import java.io.PrintWriter;
import java.math.BigDecimal;
import java.math.RoundingMode;
import java.util.Scanner;

public class Task_1263 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        String str = in.nextLine();
        int candidateCount = Integer.valueOf(str.substring(0, 1));
        int votersCount = Integer.valueOf(str.substring(2));
        int[] votersForCandidates = new int[candidateCount];
        double[] votersPercents = new double[candidateCount];
        int voter;
        for (int i = 0; i < votersCount; i++) {
            voter = Integer.valueOf(in.nextLine());
            votersForCandidates[voter - 1]++;
        }
        for (int i = 0; i < candidateCount; i++) {

            votersPercents[i] = new BigDecimal((double) 100 * votersForCandidates[i] / votersCount).setScale(2, RoundingMode.HALF_UP).doubleValue();
        }
        for (int i = 0; i < candidateCount; i++) {
            out.printf("%.2f%%%n", votersPercents[i]);
        }
        out.flush();
    }
}

42 8
2 4 6 13 14 14 19 26 26 30 60 89 89 89 89 90 90 90 90 90 91 91 93 94 94 94 94 94 94 94 95 95 95 96 96 97 98 98 98 99 100 100
184 186 209 402 408 488 621 622

$ Before going to the solution, make sure that you had tried enough, please.

$ Code:-

#include<bits/stdc++.h>
#define START int main(){
#define END system("PAUSE");return 0;}
using namespace std;

START

    int n, guest=2;
    cin >> n;
    guest += n;
    while(n--)
    {
        string s;
        cin >> s;
        for(int i=0; i<s.size(); i++)
        {
            if(s[i]=='+')
            {
                guest++;
            }
        }
    }
    if(guest != 13)
    {
        cout << 100*guest << endl;
    }
    else
    {
        cout << 100*(guest+1) << endl;
    }

END


///Thank You and Pray for me.

Edited by author 05.03.2020 17:01

I've used BST but still cannot optimize it... similar code on C is accepted with 0.171s 300kb but Python is processing the code way longer and TLE #14. And no solutions on python so far. Is there something that can be done?

The minimal output is: 1-2*5 = -9.

Edited by author 04.03.2020 22:45

consider follow string: ABZZBBBZZA.
2 polindroms: ZZBBBZZ and ABA - correctly for this problem?