Codeforces Round #392 (Div. 2)

╥﹏╥…一打cf本子就死机系列……Rating - -;
//rp并没有++;
蒟蒻默默补题解

A. Holiday Of Equality
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.

Totally in Berland there are n citizens, the welfare of each of them is estimated as the integer in ai burles (burle is the currency in Berland).

You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king’s present. The king can only give money, he hasn’t a power to take away them.

Input
The first line contains the integer n (1 ≤ n ≤ 100) — the number of citizens in the kingdom.

The second line contains n integers a1, a2, …, an, where ai (0 ≤ ai ≤ 106) — the welfare of the i-th citizen.

Output
In the only line print the integer S — the minimum number of burles which are had to spend.

Examples
input
5
0 1 2 3 4
output
10
input
5
1 1 0 1 1
output
1
input
3
1 3 1
output
4
input
1
12
output
0
Note
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.

In the second example it is enough to give one burle to the third citizen.

In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.

In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.

题目大意：补差
把整个序列变成最大数所需要的差值之和
模拟√

#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 205;
int n,num[MAXN],maxn = 0;
long long ans = 0;

int main()
{
    memset(num,0,sizeof(num));
    scanf("%d",&n);
    if(n <= 1)
    {
        puts("0");
        return 0;
    }
    for(int i = 1;i <= n; i ++)
        scanf("%d",&num[i]),maxn = max(maxn,num[i]);
        //亲测排序慢一倍
    for(int i = 1; i <= n; i ++)
        ans += maxn - num[i];
    printf("%I64d\n",ans);
    return 0;
}

B. Blown Garland
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.

Now he has a goal to replace dead light bulbs, however he doesn’t know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.

It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like “RYBGRYBGRY”, “YBGRYBGRYBG”, “BGRYB”, but can not look like “BGRYG”, “YBGRYBYGR” or “BGYBGY”. Letters denote colors: ‘R’ — red, ‘B’ — blue, ‘Y’ — yellow, ‘G’ — green.

Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.

Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:

‘R’ — the light bulb is red,
‘B’ — the light bulb is blue,
‘Y’ — the light bulb is yellow,
‘G’ — the light bulb is green,
‘!’ — the light bulb is dead.
The string s can not contain other symbols except those five which were described.

It is guaranteed that in the given string at least once there is each of four letters ‘R’, ‘B’, ‘Y’ and ‘G’.

It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line “GRBY!!!B” can not be in the input data.

Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.

Examples
input
RYBGRYBGR
output
0 0 0 0
input
!RGYB
output
0 1 0 0
input
!!!!YGRB
output
1 1 1 1
input
!GB!RG!Y!
output
2 1 1 0
Note
In the first example there are no dead light bulbs.

In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.

QAQ这个题充分说明了学好英语的重要性
QAQ给一个串，RBGY表示灯泡颜色，！表示这个灯泡坏了要换掉，整个串由四个字母重复组成，BGRYG之类的非重复序列不合法
计算需要四种颜色的灯泡各多少个

开头是可以！！！！！的
输入是可以！！！R！！！！！！Y！！G！！B！！！！！！！！的
最开始傻傻想找串匹配
(:зゝ∠)智障夏浅玖
事实上，只有四个字母组成的块重复出现，不会是RBGYRRBGYR之类的
那我当前位置是x，x - 4和x + 4的位置都一定会和x位置上的字母一样
直接暴力就可以了……反正一定得出现这个字母……
//哭晕

#include 
#include 
#include 
#include 
using namespace std;
string s;
int kr = 0,kb = 0,kg = 0,ky = 0;
int len;

void cal(int x)
{
    if(s[x] == 'R') kr ++;
    if(s[x] == 'Y') ky ++;
    if(s[x] == 'B') kb ++;
    if(s[x] == 'G') kg ++;
}

int main()
{
    cin >> s;
    len = s.length();
    for(int i = 0; i < len; i ++)
    {
        if(s[i] == '!')
        {
            if(i > 3 &&  s[i - 4] != '!')   s[i] = s[i - 4];
            else if(i < len - 4 && s[i + 4] != '!') s[i] = s[i + 4];
            else
            {
                int j = i + 4 + 4;
                while(j < len)
                {
                    if(s[j] != '!')
                    {
                        s[i] = s[j];
                        break;
                    } 
                    else j += 4;
                }
                 j = i - 4;
                while(j >= 0)
                {
                    if(s[j] != '!')
                    {
                        s[i] = s[j];
                        break;
                    } 
                    else j -= 4;
                }
            }
            cal(i);
        }
    }
    printf("%d %d %d %d\n", kr, kb, ky, kg);
    return 0;
}

C. Unfair Poll
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.

Seating in the class looks like a rectangle, where n rows with m pupils in each.

The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, …, the n - 1-st row, the n-th row, the n - 1-st row, …, the 2-nd row, the 1-st row, the 2-nd row, …

The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, …, the m-th pupil.

During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:

the maximum number of questions a particular pupil is asked,
the minimum number of questions a particular pupil is asked,
how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.

Input
The first and the only line contains five integers n, m, k, x and y (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018, 1 ≤ x ≤ n, 1 ≤ y ≤ m).

Output
Print three integers:

the maximum number of questions a particular pupil is asked,
the minimum number of questions a particular pupil is asked,
how many times the teacher asked Sergei.
Examples
input
1 3 8 1 1
output
3 2 3
input
4 2 9 4 2
output
2 1 1
input
5 5 25 4 3
output
1 1 1
input
100 100 1000000000000000000 100 100
output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:

the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;
the pupil from the first row who seats at the third table;
the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;
the pupil from the first row who seats at the third table;
the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:

the pupil from the first row who seats at the first table;
the pupil from the first row who seats at the second table;
the pupil from the second row who seats at the first table;
the pupil from the second row who seats at the second table;
the pupil from the third row who seats at the first table;
the pupil from the third row who seats at the second table;
the pupil from the fourth row who seats at the first table;
the pupil from the fourth row who seats at the second table, it means it is Sergei;
the pupil from the third row who seats at the first table;

奇怪的老师有奇怪的提问方式
从第一排提问到最后一排再提问回来
对于每一排，从第一个提问
有个小同学觉得这样不公平，于是开始算每节课老师提问最多的人和最少的人分别多少次，还有自己被提问多少次
感人的计算

注意翻译……看明白题目以后还是好打的//睁眼说瞎话

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int MAXN = 205;
ll n,m,k,x,y;//行列 问题数 他的坐标
ll maxn = 0,minn = 1e18 + 7;
ll num[MAXN][MAXN];

ll cal(int i,int j,ll n,ll m,ll k)
{
    ll cnt = 0;
    if(n == 1)  
    {
        cnt += k / m;
        if(j <= k % m)  cnt ++;
        return cnt;
    }

    ll ssr = 2 * n * m - 2 * m;
    ll hhh = (i - 1) * m + j;

    cnt += k / ssr; k %= ssr;
    if(i != 1 && i != n)cnt <<= 1;
    if(hhh <= k)    cnt ++;
    if(i == 1 || i == n) return cnt;


    i = n - i + 1;
    hhh = (i - 1) * m + j;
    k -= (n - 1) * m;
    if(hhh <= k)    cnt ++;
    return cnt;
}

int main()
{
    scanf("%I64d %I64d %I64d %I64d %I64d",&n,&m,&k,&x,&y);

    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            num[i][j] = cal(i,j,n,m,k),maxn = max(maxn,num[i][j]),minn = min(minn,num[i][j]);
    printf("%I64d %I64d %I64d\n",maxn,minn,num[x][y]);
    return 0;
} 

剩下的题看不懂&&不会做……
(:зゝ∠)蒟蒻Orz各位大大