Codeforces Round #262 (Div. 2) 题解

A. Vasya and Socks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya
takes off the used socks and throws them away. Every m-th day (at days with numbers m, 2m, 3m, ...)
mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?

Input

The single line contains two integers n and m (1 ≤ n ≤ 100; 2 ≤ m ≤ 100),
separated by a space.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input

2 2

output

input

9 3

output

传送门：点击打开链接

解题思路：

水题，简单模拟。

代码：

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long lint;

typedef double DB;

//const int MAXN = ;

int main()

{

    int n, m, t = 0;

    scanf("%d%d", &n, &m);

    while(n)

    {

        n--;

        t++;

        if(0 == t%m) n++;

    }

    printf("%d\n", t);

    return 0;

}

B. Little Dima and Equation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.

Find all integer solutions x (0 < x < 109) of
the equation:

x = b·s(x)a + c,

where a, b, c are
some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x.

The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c.
Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.

Input

The first line contains three space-separated integers: a, b, c (1 ≤ a ≤ 5; 1 ≤ b ≤ 10000; - 10000 ≤ c ≤ 10000).

Output

Print integer n — the number of the solutions that you've found. Next print n integers
in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.

Sample test(s)

input

3 2 8

output

3

10 2008 13726

input

1 2 -18

output

input

2 2 -1

output

4

1 31 337 967

传送门：点击打开链接

解题思路：

S（X）的取值从1到81。我们能够通过枚举S（x）的值得到x的值，检验是否符合。

代码：

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long lint;

typedef double DB;

const int MAX = 1e9;

const int MAXN = 100;

lint ans[100];

int fun(lint x)

{

    int ret = 0;

    while(x)

    {

        ret += x%10;

        x /= 10;

    }

    return ret;

}

int main()

{

    int a, b, c, n, m = 0;

    scanf("%d%d%d", &a, &b, &c);

    for(int i=1; i<=81; ++i)

    {

        lint x = 1ll*b*pow(i*1.0,a) + 1ll*c;

        if(x<MAX && x>0 && i==fun(x)) ans[m++] = x;

    }

    sort(ans, ans+m);

    printf("%d\n", m);

    for(int i=0; i<m; ++i)

    {

        if(i) printf(" ");

        printf("%I64d", ans[i]);

    }

    printf("\n");

    return 0;

}

C. Present

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his informatics teacher is going to have a birthday and the beaver has decided to prepare a present for her. He planted n flowers
in a row on his windowsill and started waiting for them to grow. However, after some time the beaver noticed that the flowers stopped growing. The beaver thinks it is bad manners to present little flowers. So he decided to come up with some solutions.

There are m days left to the birthday. The height of the i-th
flower (assume that the flowers in the row are numbered from 1 to n from
left to right) is equal to ai at
the moment. At each of the remaining m days the beaver can take a special watering and water w contiguous
flowers (he can do that only once at a day). At that each watered flower grows by one height unit on that day. The beaver wants the height of the smallest flower be as large as possible in the end. What maximum height of the smallest flower can he get?

Input

The first line contains space-separated integers n, m and w (1 ≤ w ≤ n ≤ 105; 1 ≤ m ≤ 105).
The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the maximum final height of the smallest flower.

Sample test(s)

input

6 2 3

2

2 2 2 1 1

output

input

2 5 1

5 8

output

传送门：点击打开链接

解题思路：

对所求解的值进行二分。

ps：这里的b数组大小是n+w。

代码：

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long lint;

typedef double DB;

const int MAXN = 2e5+10;

const int INF = 2e9;

lint a[MAXN], b[MAXN], ans;

int n, m, w;

bool check(lint k)

{

    memset(b, 0, sizeof(b));

    lint sum = 0, d = 0;

    for(int i=1; i<=n; ++i)

    {

        sum += b[i];

        lint tp = k - a[i] - sum;

        if(tp > 0)

        {

            sum += tp;

            b[i+w] -= tp;

            d += tp;

      //      printf("%I64d %I64d\n", tp, d);

            if(d > m) return false;

        }

    }

    return true;

}

int main()

{

    scanf("%d%d%d", &n, &m, &w);

    for(int i=1; i<=n; ++i)

        scanf("%I64d", a+i);

    lint l = 1, r = 1ll*INF;

    while(l <= r)

    {

        lint mid = (l+r)>>1;

        if(check(mid))

            l = mid + 1, ans = mid;

        else

            r = mid - 1;

    //    printf("%I64d %I64d\n", l, r);

    }

    printf("%I64d\n", ans);

    return 0;

}

巴特西

Codeforces Round #262 (Div. 2) 题解

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