【解题报告】AtCoder ABC115 （附英文题目）

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A - Christmas Eve Eve Eve

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Time limit : 2sec / Memory limit : 1024MB

Score : 100 points

Problem Statement

In some other world, today is December D-th.

Write a program that prints Christmas if D=25, Christmas Eve if D=24, Christmas Eve Eve if D=23 and Christmas Eve Eve Eve if D=22.

Constraints

• 22≤D≤25
• D is an integer.

签到题1，考察if语句的用法。。。

#include <iostream>
int main()
{
    int n; std::cin >> n;
    if (n == 25) puts("Christmas");
    if (n == 24) puts("Christmas Eve");
    if (n == 23) puts("Christmas Eve Eve");
    if (n == 22) puts("Christmas Eve Eve Eve");
    return 0;
}

 

B - Christmas Eve Eve

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Time limit : 2sec / Memory limit : 1024MB

Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1≤i≤N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N−1 items cost their regular prices. What is the total amount he will pay?

Constraints

• 2≤N≤10
• 100≤p_i≤10000
• p_i is an even number.

签到题2，考察for语句的用法。。。

#include <iostream>
int main()
{
    int n, x, si = 0, mi = -1; std::cin >> n;
    for (int i = 1; i <= n; ++i) 
        std::cin >> x, si += x, mi = std::max(mi, x);
    std::cout << si - (mi >> 1);
    return 0;
}

 

C - Christmas Eve

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Time limit時間制限 : 2sec / Memory limitメモリ制限 : 1024MB

Score : 300 points

Problem Statement

In some other world, today is Christmas Eve.

There are N trees planted in Mr. Takaha's garden. The height of the i-th tree (1≤i≤N) is h_i meters.

He decides to choose K trees from these trees and decorate them with electric lights. To make the scenery more beautiful, the heights of the decorated trees should be as close to each other as possible.

More specifically, let the height of the tallest decorated tree be h_max meters, and the height of the shortest decorated tree be h_min meters. The smaller the value h_max−h_min is, the better. What is the minimum possible value of h_max−h_min?

Constraints

• 2≤K<N≤10⁵
• 1≤h_i≤10⁹
• h_i is an integer.

签到题3，考察sort函数的用法。。。

#include <iostream>
#include <algorithm>
int h[101010];
int main()
{
    int n, mi = 2e9 + 1, k; std::cin >> n >> k;
    for (int i = 1; i <= n; ++i) std::cin >> h[i];
    std::sort(h + 1, h + n + 1);
    for (int i = k; i <= n; ++i) mi = std::min(mi, h[i] - h[i - k + 1]);
    std::cout << mi;
    return 0;
}

D - Christmas

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Time limit時間制限 : 2sec / Memory limitメモリ制限 : 1024MB

Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

• A level-0 burger is a patty.
• A level-L burger (L≥1) is a bun, a level-(L−1) burger, a patty, another level-(L−1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

• 1≤N≤50
• 1≤X≤( the total number of layers in a level-N burger )
• N and X are integers.

#include <iostream>
#include <algorithm>

long long b[60], h[60], n, r, ans;

void solve(int now)
{
    if (r <= 0) return;
    if (!now) { --r, ++ans; return; }
    if (r >= h[now]) { r -= h[now], ans += b[now]; return; } 
    --r; if (r > 0) solve(now - 1);
    if (r > 0) ++ans, --r;
    --r; if (r > 0) solve(now - 1);
}

int main()
{
    scanf("%lld%lld", &n, &r);
    h[0] = 1, b[0] = 1;
    for (int i = 1; i <= 50; ++i) h[i] = (h[i - 1] << 1) + 3, b[i] = 1 + (b[i - 1] << 1);
    solve(n); std::cout << ans;
    return 0;
}

太简单辣。。。。。。。。