Acmer in HDU-ACM team are ambitious, especially shǎ崽, he can spend time in Internet bar doing problems overnight. So many girls want to meet and Orz him. But Orz him is not that easy.You must solve this problem first.
The problem is :
Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
For example, give you 1 2 3 4 5, you should output 5 1 4 2 3

Input
There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.
Output
Output a sequence of distinct integers described above.

Sample Input
5
1 2 3 4 5
Sample Output
5 1 4 2 3

#include #include using namespace std; int main() { int a[10000], i, j, n; while (~scanf("%d", &n)) { for (i = 0; i < n; ++i) scanf("%d", a + i); sort(a, a + n); for (i = 0, j = n - 1; i <= (n - 1) / 2 - 1; i++, j--) { printf("%d %d ", a[j], a[i]); } if (n % 2) printf("%d\n", a[i]); else printf("%d %d\n", a[j], a[i]); } return 0; }

Calendars in widespread use today include the Gregorian calendar, which is the de facto international standard, and is used almost everywhere in the world for civil purposes. The Gregorian reform modified the Julian calendar’s scheme of leap years as follows:
Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100; the centurial years that are exactly divisible by 400 are still leap years. For example, the year 1900 is not a leap year; the year 2000 is a leap year.
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In this problem, you have been given two dates and your task is to calculate how many days are between them. Note, that leap years have unusual number of days in February.
Look at the sample to understand what borders are included in the aswer.

#include int sum(int y,int m,int d) { char x[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int i,s=0; for(i=1; is2) printf("%ld\n",s1-s2); else printf("%ld\n",s2-s1); }

You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in s. I.e. if s=“abca” then you have to press ‘a’, then ‘b’, ‘c’ and ‘a’ again.
You know that you will spend m wrong tries to perform the combo and during the i-th try you will make a mistake right after pi-th button (1≤pi I.e. if s=“abca”, m=2 and p=[1,3] then the sequence of pressed buttons will be ‘a’ (here you’re making a mistake and start performing the combo from the beginning), ‘a’, ‘b’, ‘c’, (here you’re making a mistake and start performing the combo from the beginning), ‘a’ (note that at this point you will not perform the combo because of the mistake), ‘b’, ‘c’, ‘a’.
Your task is to calculate for each button (letter) the number of times you’ll press it.
You have to answer t independent test cases.

The first line of the input contains one integer t (1≤t≤104) — the number of test cases.
Then t test cases follow.
The first line of each test case contains two integers n and m (2≤n≤2?105, 1≤m≤2?105) — the length of s and the number of tries correspondingly.
The second line of each test case contains the string s consisting of n lowercase Latin letters.
The third line of each test case contains m integers p1,p2,…,pm (1≤pi It is guaranteed that the sum of n and the sum of m both does not exceed 2?105 (∑n≤2?105, ∑m≤2?105).
It is guaranteed that the answer for each letter does not exceed 2?109.

For each test case, print the answer — 26 integers: the number of times you press the button ‘a’, the number of times you press the button ‘b’, …, the number of times you press the button ‘z’.

Input
3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4
Output
4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2

The first test case is described in the problem statement. Wrong tries are “a”, “abc” and the final try is “abca”. The number of times you press ‘a’ is 4, ‘b’ is 2 and ‘c’ is 2.
In the second test case, there are five wrong tries: “co”, “codeforc”, “cod”, “co”, “codeforce” and the final try is “codeforces”. The number of times you press ‘c’ is 9, ‘d’ is 4, ‘e’ is 5, ‘f’ is 3, ‘o’ is 9, ‘r’ is 3 and ‘s’ is 1.

#include using namespace std; const int maxn = 2e5 + 10; char s[maxn]; int a[maxn]; int cnt[26]; int main() { int i, t, n, m, tmp, cur; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &m); scanf("%s", s + 1); for(i = 1; i <= n; i++) a[i] = 0; for(i = 0; i < 26; i++) cnt[i] = 0; for(i = 0; i < m; i++) { scanf("%d", &tmp); a[tmp]++; } for(i = 1, cur = m + 1; i <= n; i++) { cnt[s[i] - 'a'] += cur, cur -= a[i]; } for(i = 0; i < 25; i++) printf("%d ", cnt[i]); printf("%d", cnt[25]); if(t) printf("\n"); } return 0; }

Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings “pop”, “noon”, “x”, and “kkkkkk” are palindromes, while strings “moon”, “tv”, and “abab” are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.

The first line contains two integers n and m (1≤n≤100, 1≤m≤50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.

In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don’t print this line at all.

Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba

In the first example, “battab” is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.

在这里插入代#include #include using namespace std; char s[105][55]; int mp[1005],a[1005]; int m; int check(int i,int j) { for(int k=0; k=2; i--) { if(i%2==0) printf("%s",s[a[i]]); } printf("\n"); return 0; } 码片

After your death, you’re sent to a mysterious room. There are two guardian cats and two doors, one goes to heaven of AC problems and another goes to hell NO. One cat likes all divisors of an integer a and the other cat likes all multiples of an integer b. You where asked to write all integers that satisfies both. If you answer correctly, you may go to heaven of AC problems, full of balloons, otherwise you’re sent to hell NO.

The input consists of two integers a and b, where 1?≤?b?≤?a?≤?1012.

The output consists of one line with all integers from smallest to largest that satisfies both guardian cats.

Input
12 3
Output
3 6 12
Input
10 3
Output
Input
128 2
Output
2 4 8 16 32 64 128

#include #include #define ll long long using namespace std; int main() { ll i,j=0,a,b,c[100005]; scanf("%lld%lld",&a,&b); for(i=1; i*i<=a; i++) { if(a%i==0) { c[j++]=i; if(i!=a/i) c[j++]=a/i; } } sort(c,c+j); for(i=0; i