Codeforces Round #196 (Div. 2)

涨了那么一点 。。

A纯水 把CF都交挂了。。

B 扩展与x相同 或与y相同

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 int gcd(int a,int  b)
 8 {
 9     return b==0? a:gcd(b,a%b);
10 }
11 int main()
12 {
13     int i,j,a,b,c,d;
14     cin>>a>>b>>c>>d;
15     int x =a*c/gcd(a,c);
16     int y = b*d/gcd(b,d);
17     int x1 = b*(x/a);
18     int y1 = d*(x/c);
19     int x2 = a*(y/b);
20     int y2 = c*(y/d);
21     if(x1>=y1)
22     {
23 
24         y1 = x1-y1;
25         int o = gcd(x1,y1);
26         printf("%d/%d\n",y1/o,x1/o);
27     }
28     else
29     {
30        y2 = x2-y2;
31        int o = gcd(x2,y2);
32        printf("%d/%d\n",y2/o,x2/o);
33     }
34     return 0;
35 }
View Code

C 题 搞笑。。我把2倍看成了平方 然后WA到结束。。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define LL __int64
 8 #define mod 1000000009
 9 LL poww(LL a,LL n)
10 {
11     LL t;
12     if(n==0)  return 1%mod;
13     if(n==1)  return a%mod;
14     t=poww(a,n/2);
15     t=t*t%mod;
16     if((n&1)==1) t=t*a%mod;
17     return t;
18 }
19 int main()
20 {
21     LL i,k,n,m;
22     cin>>n>>m>>k;
23     LL o = n-m;
24     if((o+1)*(k-1)>=m)
25     {
26         printf("%I64d\n",m);
27     }
28     else
29     {
30         LL s = o*(k-1);
31         LL y = (m-s)/k;
32         LL ans = (k*poww(2,y+1)%mod-2*k%mod)%mod;
33         ans = (ans+(m-s-y*k)%mod+s)%mod;
34         if(ans<0)
35         ans+=mod;
36         printf("%I64d\n",ans);
37     }
38     return 0;
39 }
View Code

D题 找m点里面最长的两个点 再以这两个点进行两遍dfs 如果某点离这个两个点的距离都小于d 那么其它的也小于

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define N 100010
 8 struct node
 9 {
10     int u,v,next;
11 }ed[N<<1];
12 int t,head[N];
13 void init()
14 {
15     t =0;
16     memset(head,-1,sizeof(head));
17 }
18 void add(int u,int v)
19 {
20     ed[t].u = u;
21     ed[t].v = v;
22     ed[t].next = head[u];
23     head[u] = t++;
24 }
25 void dfs(int u,int pre,int dd,int o[])
26 {
27     o[u] = dd;
28     for(int i = head[u] ; i != -1; i = ed[i].next)
29     {
30         int v = ed[i].v;
31         if(v==pre)
32         continue;
33         dfs(v,u,dd+1,o);
34     }
35 }
36 int s1[N],s2[N],s3[N],f[N];
37 int main()
38 {
39     int i,n,m,d,k;
40     init();
41     cin>>n>>m>>d;
42     for(i =1; i <= m ; i++)
43     scanf("%d",&f[i]);
44     for(i = 1; i < n ;i++)
45     {
46         int u,v;
47         scanf("%d%d",&u,&v);
48         add(u,v);
49         add(v,u);
50     }
51     dfs(f[1],-1,0,s1);
52     k = f[1];
53     for(i = 2; i <= m ; i++)
54     {
55         if(s1[k]<s1[f[i]])
56         k = f[i];
57     }
58     dfs(k,-1,0,s2);
59     k = f[1];
60     for(i = 2; i <= m ; i++)
61     if(s2[k]<s2[f[i]])
62         k = f[i];
63     dfs(k,-1,0,s3);
64     int ans = 0;
65     for(i = 1; i <= n ; i++)
66         if(s2[i]<=d&&s3[i]<=d)
67             ans++;
68     printf("%d\n",ans);
69     return 0;
70 
71 }
View Code

 

posted @ 2013-08-17 19:48  _雨  阅读(228)  评论(0编辑  收藏  举报