First, let give the bits some names, I put them inside the parentheses:
Petal 1 (p1) = II (c2) + III (c3) + IV (c4)
Petal 2 (p2) = I (c1) + III (c3) + IV (c4)
Petal 3 (p3) = I (c1) + II (c2) + IV (c4)

As p1, p2 and p3 are created from c1, c2, c3, c4, we check each change in c1, c2, c3, c4:
When c1 change: p2 and p3 won't match the value calculated using the formular above (i.e: c1 + c3 + c4 != p2 and c1 + c2 + c4 != p3)
When c2 change: p1 and p3 will go wrong
When c3 change: p1 and p2 will go wrong
As c4 contribute to the value of all three petals, if it is changed then p1, p2 and p3 will all go wrong.

In short:
Calculate ep1, ep2 and ep3 from the first 4 bits (the e in ep stand for expected), then compare them with the given p1, p2 and p3:
    All three pairs do not match: c4 is changed
    p1 and p2 do not match:       c3 is changed
    p1 and p3 do not match:       c2 is changed
    p2 and p3 do not match:       c1 is changed
    Only p1 not match:            p1 is changed
    Only p2 not match:            p2 is changed
    Only p3 not match:            p3 is changed

The problem is rather easy, just compare squares due to rationality of numbers

I print 2 * (n - 1) when n <= m and 2 * (m - 1) + 1 when n > m but got WA on test 13. What's wrong with my solution?

Oh, I forgot to delete my check for m = 1 (in this case I output 1). This test seem to be
1 1
0

Edited by author 03.10.2018 17:42

For those looking for a hint: I use stack.
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Push the characters into stack, when a non-alphabetic character is read, pop out all the characters in stack and print them before printing the non-alphabetic character.

Edited by author 02.10.2018 15:04

#include <stdio.h>

int main()
{
    char a[3],b[3];
    gets(a);
    gets(b);
    int x;
    if (a[3]=='0'||a[3]=='2'||a[3]=='4'||a[3]=='6'||a[3]=='8')
        x=1;
    if (b[3]=='1'||b[3]=='3'||b[3]=='5'||b[3]=='7'||b[3]=='9')
        x=1;
   if(x==1) printf("yes");
     else printf("no");
}

Do I uderstand it right, that numbers are placed
in increasing order, there's nothing said about it??? Правильно понимаю, что числа должны быть в порядке возрастания?

You can solve this problem quickly by using the multiset and set data structures. To find the answer, use the set iterator.

#! /usr/bin/env python
# -*- coding: utf-8 -*-
from itertools import count


mailBox = {"Alice": 1, "Ariel": 1, "Aurora": 1,"Phil": 1, "Peter": 1,"Olaf": 1,"Phoebus": 1, "Ralph": 1, "Robin": 1,
           "Bambi": 2, "Belle": 2,"Bolt": 2, "Mulan": 2,"Mowgli": 2,"Mickey": 2,"Silver": 2,"Simba": 2,"Stitch": 2,
           "Dumbo": 3, "Genie": 3,"Jiminy": 3, "Kyzko": 3,"Kida": 3,"Kenai": 3,"Tarzan": 3,"Tiana": 3,"Winnie": 3,
           }

# кол-во писем
n = int(raw_input())
# list с именами адресатов
inputMailFrom = []
# цикл по заполению адресатов
for n in range(n):
    inputMailFrom.append(mailBox.get(raw_input()))
step = []


if len(inputMailFrom) == 1:
    if  inputMailFrom[0] == 1:
        print (0)
    if inputMailFrom[0] == 2:
        print (1)
    if inputMailFrom[0] == 3:
        print (2)
else:
    for i in range(n):
        step.append(abs(inputMailFrom[i]-inputMailFrom[i+1]))
    print (sum(step)

n=int(input())
s1=['Alice','Ariel','Aurora','Phil','Peter','Olaf','Phoebus','Ralph','Robin' ]
s2=['Bambi','Belle', 'Bolt','Mulan','Mowgli','Mickey','Silver','Simba','Stitch']
s3=['Dumbo','Genie','Jiminy','Kuzko','Kida','Kenai','Tarzan','Tiana','Winnie']
a=s1
k=0
for i in range(n):
    s=input()
    if s in s1:
        if a==s1:
            continue
        elif a==s2:
            k+=1
            a=s1
        elif a==s3:
            k+=2
            a=s1
    elif s in s2:
        if a==s1:
            k+=1
            a=s2
        elif a==s2:
            continue
        elif a==s3:
            k+=1
            a=s2
    elif s in s3:
        if a==s1:
            k+=2
            a=s3
        elif a==s2:
            k+=1
            a=s2
        elif a==s3:
            continue
print(k)

give me tests,please! all my tests give ok results. i have wa 19. help!

What answer in this case?

Нужно убрать максимальное количество ребёр, но чтобы все вершины имели ненулевую степень.

I thought that stack is the best thing in this case but I've seen 0.015 and 0.001 sec . I've got 0.046

First method:
Use multiple baskets, each basket is a list (in my case, i use vector<int>). Put the date d into the basket numbered d % number_of_baskets. To check the student answer, just pick the basket out and perform binary search. I chose 100 000 as the number of baskets, tried 1 000 000 baskets but it works much worse.

Second method:
Use vector<bool> (not bool array, the later cost 8 times more) to mark numbers under 300 000 000 (actually the vector size can go up to 500 000 000, but for some unknown reason, this smaller size run faster). Create another vector<int> to store the numbers that can't be marked using the bool vector above. Use binary_search to check for existence.

It seem like the IO is bottleneck, I used ios::sync_with_stdio(false) and my program run from ~1.5s to ~0.2s

Edited by author 01.10.2018 12:46

seems I have solved memory problem..  but I don't know if bitset can fit in time

can O(1e8) fit into 1 sec??

Edited by author 27.09.2018 06:21

If you got WA, especially if you got WA9
First line is input string, second is answer

TEST. TEST? TEST! TEST.
Test. Test? Test! Test.

??????????????????????
??????????????????????

!TEST!TEST!
!Test!Test!

A
A

TEST
    TEST
        TEST
Test
    test
        test

TEST     T      TEST
Test     t      test

Also try empty input.

Edited by author 30.09.2018 19:14

This enchanted test 8: I just can not get through it. Maybe there is just that unrepresentable case of "impossible", the emergence of which, under given conditions, I can not even imagine?

I dont understand: when I had swaped 'Impossible' to '0' my program solved more tests than late.

void fallBack(int angle, int M, int curEnergy, int curAmmo);
void advance(int angle, int M, int curEnergy, int curAmmo);
int findX(int curEnergy);

int main()
{
    int curEnergy, curAmmo;
    char behaviour;
    int M, MP, angle;
    int N, NP, offset;
    scanf_s("%i %i", &curEnergy, &curAmmo);
    scanf_s(" %c", &behaviour);
    scanf_s("%i %i %i", &M, &MP, &angle);
    if (behaviour == 'A') {
        scanf_s("%i %i", &N, &NP);
    }
    else if (behaviour == 'P') {
        scanf_s("%i", &offset);
        offset *= -1;

    }
    while (1) {
        switch (behaviour) {
        case 'A':
            if ((N*NP) > (M*MP * 3)) {
                advance(angle, M, curEnergy, curAmmo);
                return 0;
            }
            else {
                fallBack(angle, M, curEnergy, curAmmo);
                return 0;
            }
        case 'D':
            if ((M * 20) >= curAmmo) {
                fallBack(angle, M, curEnergy, curAmmo);
                return 0;
            }
            else {
                behaviour = 'G';
                break;
            }
        case 'G':
            if (M == 0) {
                printf("STOP");
            }
            if (abs(angle) < 5) {
                int P = (int)fmin(20, curAmmo);
                printf("FIRE %i", P);
                return 0;
            }
            else if (angle >= 5) {
                printf("LEFT %i", findX(curEnergy));
                return 0;
            }
            else if (angle <= -5) {

                printf("RIGHT %i", findX(curEnergy));
                return 0;
            }
        case 'P':
            if (M != 0) {
                behaviour = 'D';
                break;
            }
            else {
                if (offset >= -180 && offset <= -160) {
                    printf("BACKWARD %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > -160 && offset < -90) {
                    printf("RIGHT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= -90 && offset < -20) {
                    printf("LEFT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= -20 && offset <= 20) {
                    printf("FRONT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > 20 && offset <= 90) {
                    printf("RIGHT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > 90 && offset < 160) {
                    printf("LEFT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= 160 && offset <= 180) {
                    printf("BACKWARD %i", findX(curEnergy));
                    return 0;
                }
            }
        }
    }


    return 0;
}

void fallBack(int angle, int M, int curEnergy, int curAmmo)
{
    if (abs(angle) >= 5 || M == 0) {
        printf("BACKWARD %i", findX(curEnergy));
    }
    else {
        int P = (int)fmin(20, curAmmo);
        printf("FIRE %i", P);
    }
}
void advance(int angle, int M, int curEnergy, int curAmmo)
{
    if (abs(angle) >= 10 || M == 0) {
        printf("FRONT %i", findX(curEnergy));
    }
    else {
        int P = (int)fmin(20, curAmmo);
        printf("FIRE %i", P);
    }
}
int findX(int curEnergy)
{
    int X;
    if (curEnergy > 100) {
        X = 100;
    }
    else {
        X = curEnergy;
    }
    return X;
}