codeforces Round #441 div2

A. Trip For Meal
time limit per test 1 second
memory limit per test 512 megabytes
input standard input
output standard output
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit’s and Owl’s houses is a meters, between Rabbit’s and Eeyore’s house is b meters, between Owl’s and Eeyore’s house is c meters.

For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit’s house and has a meal for the first time. Each time when in the friend’s house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend’s house, the supply of honey in other friend’s houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit’s and Owl’s houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit’s and Eeyore’s houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl’s and Eeyore’s houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
input
3
2
3
1
output
3
input
1
2
3
5
output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit’s house, then in Owl’s house, then in Eeyore’s house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit’s house and that is for him. So he doesn’t have to walk anywhere at all.

题意：有ROE三个点，每个点都可以吃东西，RO相距a米，RE相距b米，OE相距c米，小熊要吃n次东西，最少要走多少距离，现已知小熊从R点出发（此时已经吃完了第一顿饭），R点吃完后小熊离开R点后还可以再来吃
思路：第一顿在R吃，第二顿在O或E吃（取决于哪个更近），之后就徘徊于最短边两点之间吃东西，才是最佳的

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int main(int argc,char *argv[]) {    int n,a,b,c;    scanf("%d%d%d%d",&n,&a,&b,&c);    if(n == 1) printf("%d\n",0);    else if(n == 2) printf("%d\n",min(a,b));    else printf("%d\n",min(a,b) + min(min(a,b) , c) * (n - 2));    return 0;}

B. Divisiblity of Differences
time limit per test 1 second
memory limit per test 512 megabytes
input standard input
output standard output
You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible.

Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.
Input
First line contains three integers n, k and m (2 ≤ k ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains n integers a1, a2, …, an (0 ≤ ai ≤ 109) — the numbers in the multiset.
Output
If it is not possible to select k numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print k integers b1, b2, …, bk — the selected numbers. If there are multiple possible solutions, print any of them.
Examples
input
3 2 3
1 8 4
output
Yes
1 4
input
3 3 3
1 8 4
output
No
input
4 3 5
2 7 7 7
output
Yes
2 7 7

题意：在n个数中取出k个数，使得这k个数两两之间的差都是m的倍数。
对每个数%m，余数相同那么它们的差就是m的倍数

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#define MAXN 200200using namespace std;int k,m,n,num,q[MAXN],a[MAXN],sum[MAXN],s;int main(int argc,char *argv[]) {    scanf("%d%d%d",&n,&k,&m);    for(int i=1; i<=n; ++i) scanf("%d",&a[i]),q[i] = a[i] % m;    for(int i=1; i<=n; ++i) ++sum[q[i]];    bool flag = false;    for(int i=0; i<m; ++i)        if(sum[i] >= k){ flag = true,num = i; break; }    if(flag) {        puts("Yes");        for(int i=1; i<=n; ++i)            if(q[i] == num){                if(++s == k) { printf("%d\n",a[i]); break; }                else printf("%d ",a[i]);            }    }    else puts("No");    return 0;}

C. Classroom Watch
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number x written in decimal numeral system.

Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.

Input
The first line contains integer n (1 ≤ n ≤ 109).

Output
In the first line print one integer k — number of different values of x satisfying the condition.

In next k lines print these values in ascending order.

Examples
input
21
output
1
15
input
20
output
0
Note
In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.

In the second test case there are no such x.
题目大意：题目的范围是1000000000，每个数字加起来的和最大的数是99999999，可以直接从n-81的地方开始枚举

#include<iostream>#include<vector>using namespace std;int main() {    int n;    cin >> n;    vector<int> vec;    for(int i=81; i>=0; i--) {        if(n-i < 0) continue;        int res = n - i;        int tmp = res,ans = 0;        while(tmp) {            ans += tmp % 10;            tmp /= 10;        }        if(ans + res == n) vec.push_back(res);    }    cout << vec.size() << endl;    for(int i=0; i<vec.size(); i++) cout << vec[i] << ' ';}

D. Sorting the Coins
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
He looks through all the coins from left to right;
If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 ≤ n ≤ 300 000) — number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, …, pn (1 ≤ pi ≤ n) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, …, an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.

Examples
input
4
1 3 4 2
output
1 2 3 2 1
input
8
6 8 3 4 7 2 1 5
output
1 2 2 3 4 3 4 5 1
Note
Let’s denote as O coin out of circulation, and as X — coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won’t make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO  →  OOOX
After replacement of the third coin, Dima’s actions look this way:
XOXO  →  OXOX  →  OOXX
After replacement of the fourth coin, Dima’s actions look this way:
XOXX  →  OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.

题意：有n个硬币，刚开始所有的硬币均为不流通，然后分开n次使得n个硬币流通，小明每次只能从左到右看，不能返回，且不能用手改变位置，若看的位置是流通硬币且下一个硬币是不流通的，则小明则在心中将这两个硬币交换位置，若前后两个硬币均为流通或不流通则不交换，问每次将此时流通的硬币全部排到最后的难度是多少（因为每次看完要记住当前的位置，所以看的次数越多越难，不用看时的难度为1多看一次加一）。
思路：已经在最后的相连的流通硬币是不会交换的，然后每次将一个流通硬币换到下一个流通硬币时这个流通硬币就不能再换的，要开始换下一个流通硬币，等同于就是每次看把最后的没有在尾部的流通硬币移到尾部，即每次看的次数就是没有连接在尾部的流通硬币个数。难度+1

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;typedef long long LL;const int MAXN = 300000 + 10;LL n,p[MAXN],sum,vis[MAXN],last;int main(int argc,char *argv[]) {    scanf("%lld",&n);    for(int i=0; i<n; ++i) scanf("%lld",&p[i]);    last = n,sum = 0;    printf("1");    for(int i=0; i<n; ++i) {        vis[p[i]] = true; sum ++ ;        while(vis[last]) sum--,last--;        printf(" %lld",sum + 1);    }    printf("\n");    return 0;}