AIM Tech Round (Div. 2)——ABCD

http://codeforces.com/contest/624

A.python是用来写div2的AB题的...

 a, b, x, y = map(float, raw_input().split())
 print ((b - a) / (x + y))

B.思路很简单，但你的实现一样简短吗

 n = input()
 a = sorted(map(int, raw_input().split()))
 s = 0
 x = 10 ** 9
 while a and x:
     x = min(x, a.pop())
     s += x
     x -= 1
 print s

C.注意到只有abc三种字符

b是和abc都相邻的！就很ok！

跟n-1个都相邻的肯定涂b

然后随便选个连b的涂成a再dfs

剩下的涂c最后检验就行

 import java.util.Scanner;

 public class C {
     static int[][] g = new int[505][505];
     static char[] s = new char[505];
     static int n;
     static void dfs(int x) {
         s[x] = 'a';
         for(int i = 1;i <= n;i ++)
             if(g[x][i] == 1 && s[i] == 0)
                 dfs(i);
     }
     public static void main(String []args) {
         Scanner cin = new Scanner(System.in);
         n = cin.nextInt();
         int m = cin.nextInt();
         int[] u = new int[130005];
         int[] v = new int[130005];
         int[] c = new int[505];
         for(int i = 1;i <= m;i ++) {
             u[i] = cin.nextInt();
             v[i] = cin.nextInt();
             g[u[i]][v[i]] = 1;
             g[v[i]][u[i]] = 1;
             c[u[i]] ++;
             c[v[i]] ++;
         }
         boolean ok = false;
         for(int i = 1;i <= n;i ++)
             if(c[i] == n - 1) {
                 s[i] = 'b';
                 ok = true;
             }
         if(!ok) dfs(1);
         else {
             for(int i = 1;i <= n;i ++) {
                 if(s[u[i]] == s[v[i]] && s[u[i]] == 'b') continue;
                 if(s[u[i]] == 'b' || s[v[i]] == 'b') {
                     if(s[u[i]] == 'b') dfs(v[i]);
                     else dfs(u[i]);
                     break;
                 }
             }
         }
         for(int i = 1;i <= n;i ++)
             if(s[i] == 0)
                 s[i] = 'c';
         for(int i = 1;i <= n;i ++)
             for(int j = i + 1;j <= n;j ++) {
                 boolean x = (g[i][j] != 0);
                 boolean y = (s[i] == s[j] || s[i] - s[j] == 1 || s[j] - s[i] == 1);
                 if(x ^ y) {
                     System.out.println("No");
                     System.exit(0);
                 }
             }
         System.out.println("Yes");
         for(int i = 1;i <= n;i ++)
             System.out.print(s[i]);
     }
 }

注意也可能没有b，就选第一个点涂a，dfs下去就行了

D.显然a1,a1 + 1, a1 - 1, an, an + 1, an - 1

这六个数在最后的数列里至少出现了一个

最后数列的gcd肯定在这个数里...然后枚举这个gcd

开始DP，f[0/1/2][i]代表到第i个位置还没开始remove/正在remove/已经结束remove的最小代价

状态说清了，转移方程想一想咯

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 typedef long long ll;

 int n, a, b, c[maxn], v[maxn];

 ll f[][maxn], ans = 1e18;

 void dp(int x) {
     memset(f, 0x3f3f3f3f, sizeof f);
     f[][] = f[][] = f[][] = ;
     for(int i = ;i <= n;i ++) {
         if(c[i] % x ==  || c[i] % x ==  || c[i] % x == x - ) {
             f[][i] = f[][i - ] + b * (c[i] % x != );
             f[][i] = f[][i - ] + b * (c[i] % x != );
         }
         f[][i] = min(f[][i - ], f[][i - ]) + a;
         f[][i] = min(f[][i], f[][i]);
         if(f[][i] >= ans && f[][i] >= ans) return;
     }
     ans = min(min(f[][n], f[][n]), ans);
 }

 void solve(int x) {
     for(int i = ;i * i <= x;i ++)
         if(x % i == ) {
             if(!v[i]) dp(i);
             v[i] = ;
             while(x % i == ) x /= i;
         }
     if(x != ) dp(x);
 }

 int main() {
     scanf("%d %d %d", &n, &a, &b);
     for(int i = ;i <= n;i ++)
         scanf("%d", &c[i]);
     solve(c[]), solve(c[] - ), solve(c[] + );
     solve(c[n]), solve(c[n] - ), solve(c[n] + );
     printf("%lld\n", ans);
     return ;
 }