AIM Tech Round (Div. 2)——ABCD

http://codeforces.com/contest/624

 

A.python是用来写div2的AB题的...

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

 

B.思路很简单,但你的实现一样简短吗

1 n = input()
2 a = sorted(map(int, raw_input().split()))
3 s = 0
4 x = 10 ** 9
5 while a and x:
6     x = min(x, a.pop())
7     s += x
8     x -= 1
9 print s
View Code

 

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

 1 import java.util.Scanner;
 2 
 3 public class C {
 4     static int[][] g = new int[505][505];
 5     static char[] s = new char[505]; 
 6     static int n;
 7     static void dfs(int x) {
 8         s[x] = 'a';
 9         for(int i = 1;i <= n;i ++)
10             if(g[x][i] == 1 && s[i] == 0)
11                 dfs(i);
12     }
13     public static void main(String []args) {
14         Scanner cin = new Scanner(System.in);
15         n = cin.nextInt();
16         int m = cin.nextInt();
17         int[] u = new int[130005];
18         int[] v = new int[130005];
19         int[] c = new int[505];
20         for(int i = 1;i <= m;i ++) {
21             u[i] = cin.nextInt();
22             v[i] = cin.nextInt();
23             g[u[i]][v[i]] = 1;
24             g[v[i]][u[i]] = 1;
25             c[u[i]] ++;
26             c[v[i]] ++;
27         }
28         boolean ok = false;
29         for(int i = 1;i <= n;i ++)
30             if(c[i] == n - 1) {
31                 s[i] = 'b';
32                 ok = true;
33             }
34         if(!ok) dfs(1);
35         else {
36             for(int i = 1;i <= n;i ++) {
37                 if(s[u[i]] == s[v[i]] && s[u[i]] == 'b') continue;
38                 if(s[u[i]] == 'b' || s[v[i]] == 'b') {
39                     if(s[u[i]] == 'b') dfs(v[i]);
40                     else dfs(u[i]);
41                     break;
42                 }
43             }
44         }
45         for(int i = 1;i <= n;i ++)
46             if(s[i] == 0)
47                 s[i] = 'c';
48         for(int i = 1;i <= n;i ++)
49             for(int j = i + 1;j <= n;j ++) {
50                 boolean x = (g[i][j] != 0);
51                 boolean y = (s[i] == s[j] || s[i] - s[j] == 1 || s[j] - s[i] == 1);
52                 if(x ^ y) {
53                     System.out.println("No");
54                     System.exit(0);
55                 }
56             }
57         System.out.println("Yes");
58         for(int i = 1;i <= n;i ++)
59             System.out.print(s[i]);
60     }
61 }
View Code

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

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 1000010;
 6 
 7 typedef long long ll;
 8 
 9 int n, a, b, c[maxn], v[maxn];
10 
11 ll f[3][maxn], ans = 1e18;
12 
13 void dp(int x) {
14     memset(f, 0x3f3f3f3f, sizeof f);
15     f[0][0] = f[1][0] = f[2][0] = 0;
16     for(int i = 1;i <= n;i ++) {
17         if(c[i] % x == 0 || c[i] % x == 1 || c[i] % x == x - 1) {
18             f[0][i] = f[0][i - 1] + b * (c[i] % x != 0);
19             f[1][i] = f[1][i - 1] + b * (c[i] % x != 0);
20         }
21         f[2][i] = min(f[2][i - 1], f[0][i - 1]) + a;
22         f[1][i] = min(f[1][i], f[2][i]);
23         if(f[0][i] >= ans && f[1][i] >= ans) return;
24     }
25     ans = min(min(f[0][n], f[1][n]), ans);
26 }
27 
28 void solve(int x) {
29     for(int i = 2;i * i <= x;i ++)
30         if(x % i == 0) {
31             if(!v[i]) dp(i);
32             v[i] = 1;
33             while(x % i == 0) x /= i;
34         }
35     if(x != 1) dp(x);
36 }
37 
38 int main() {
39     scanf("%d %d %d", &n, &a, &b);
40     for(int i = 1;i <= n;i ++)
41         scanf("%d", &c[i]);
42     solve(c[1]), solve(c[1] - 1), solve(c[1] + 1);
43     solve(c[n]), solve(c[n] - 1), solve(c[n] + 1);
44     printf("%lld\n", ans);
45     return 0;
46 }
View Code
posted @ 2017-08-01 22:40  ztztyyy  阅读(150)  评论(0编辑  收藏  举报