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 }

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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 }

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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 }

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