AtCoder Grand Contest 004

[TakefumiYamamura]

http://agc004.contest.atcoder.jp/

[TakefumiYamamura]

A - Divide a Cuboid

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:

There is at least one red block and at least one blue block. The union of all red blocks forms a rectangular parallelepiped. The union of all blue blocks forms a rectangular parallelepiped. Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.

Constraints

2≤A,B,C≤109

#include <iostream>
using namespace std;

long long min(long long a, long long b){
    if(a > b){
        return b;
    }else{
        return a;
    }
}

int main(){
    long long a, b, c;
    cin >> a >> b >> c ;
    if(a%2 == 0 || b%2 == 0 || c%2 == 0){
        cout << 0 << endl;
        return 0;
    }

    long long ans = min(a*b, min(b*c, c*a));
    cout << ans << endl;

}

[TakefumiYamamura]

B - Colorful Slimes

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him ai seconds.

Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N−1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.

Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

2≤N≤2,000 ai are integers. 1≤ai≤109 x is an integer. 1≤x≤109

Note

• In this problem, I have to use the dynamic programming. If I use it, I can reduce the computation time from O(N^3) to O(N^2). To more detail, if I use k times magic, I can get i-th slime min(a[i-k], ....a[i]) as bi. Also I can find that bi = min(bi, a[(i-k+N) % N] )
• Secondly I notice the work of type long long. If I set int x and int y, the statement "long long sum = x + y" cannot be work well. So I have to use long long each of x or y.

#include <iostream>
#include <algorithm>
#define ll long long
#define INF 1LL << 60
using namespace std;

int N;
ll x;
int a[2010];
int b[2010];

int main(){
  cin >> N >> x;
  for (int i = 0; i < N; ++i)
  {
    cin >> a[i];
    b[i] = a[i];
  }

  ll ans = INF;

  for (int k = 0; k < N; ++k)
  {
    ll tmp = k * x;
    for (int i = 0; i < N; ++i)
    {
      b[i] = min(b[i], a[(i - k + N) % N]);
      tmp += b[i];
    }
    ans = min(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}