NOIP2013Day1解题报告

  本来今天晚上拿13年NOIP的题目来做一下,测测能够得多少分,结果一晚上把Day1写完竟然AK了,吼吼吼

 

  D1T1,题目:http://codevs.cn/problem/3285/

  很水的一道快速幂啊,题目竟然还刚好有0号节点来降低难度,记住对于转圈之后的位置要往取模方面想

  ans=(x+m*10k)%n,原本是x号位置,每次往前走m个位置,共走了10k,最后再对于圈的大小取模即为最终位置

  模意义下可变形,ans=x+m*10k%n=x+m*(10k%n)%n

  其中10k%n不就是个快速幂取模吗?直接上模板:http://www.cnblogs.com/hadilo/p/5719139.html

  简单AC

 1 // Copyright(c) Hadilo.All Rights Reserved.
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<iostream>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<stack>
10 #include<ctime>
11 #define fre(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
12 using namespace std;
13 typedef long long LL;
14 typedef double db;
15 const long double CPS=CLOCKS_PER_SEC,TL=0.998;
16 const int oo=2147483647;
17 inline int read()
18 {
19     int re=0;
20     bool fu=0;
21     char ch=getchar();
22     while ((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
23     if (ch=='-')
24     {
25         fu=1;
26         ch=getchar();
27     }
28     while (ch>='0'&&ch<='9')
29     {
30         re=re*10+ch-'0';
31         ch=getchar();
32     }
33     return fu?-re:re;
34 }
35 int n;
36 inline int mi(LL a,int b)
37 {
38     LL k=1;
39     a%=n;
40     while (b)
41     {
42         if (b&1) k=k*a%n;
43         b>>=1;
44         a=a*a%n;
45     }
46     return (int)k;
47 }
48 int main()
49 {
50     fre("circle");
51     n=read();
52     cout<<(read()*mi(10,read())+read())%n<<endl;
53     return 0;
54 }

 

  D1T2,题目:http://codevs.cn/problem/3286/

  其实看完题目我是先写的T3再写的T2,因为感觉这个不是太好证明

  可伪证一下,把两个数组排序后对应位置即为最终答案,因为找不出反例而且感觉很正确

  则对于两个数组a、b分别离散化一下,再记一个数组c[i]为b[i]离散化的值在a[i]中的位置

  最后对 c 数组求逆序对即可,归并排序或树状数组都可以,我写的是归并排序 

 1 // Copyright(c) Hadilo.All Rights Reserved.
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<iostream>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<stack>
10 #include<ctime>
11 #define fre(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
12 using namespace std;
13 typedef long long LL;
14 typedef double db;
15 const long double CPS=CLOCKS_PER_SEC,TL=0.998;
16 const int oo=2147483647,N=100001,mo=99999997;
17 inline int read()
18 {
19     int re=0;
20     bool fu=0;
21     char ch=getchar();
22     while ((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
23     if (ch=='-')
24     {
25         fu=1;
26         ch=getchar();
27     }
28     while (ch>='0'&&ch<='9')
29     {
30         re=re*10+ch-'0';
31         ch=getchar();
32     }
33     return fu?-re:re;
34 }
35 struct gap
36 {
37     int x,num;
38 };
39 gap a[N];
40 int b[N],c[N],p[N],ans;
41 inline void gb(int l,int r)
42 {
43     if (l==r) return;
44     int mid=(l+r)>>1,i,j,top=0;
45     gb(l,mid);
46     gb(mid+1,r);
47     for (i=l,j=mid+1;i<=mid&&j<=r;)
48     {
49         if (c[i]<=c[j]) b[++top]=c[i++];
50         else
51         {
52             b[++top]=c[j++];
53             if ((ans+=mid-i+1)>mo) ans-=mo;
54         }
55     }
56     while (i<=mid) b[++top]=c[i++];
57     while (j<=r) b[++top]=c[j++];
58     for (;top;top--,r--) c[r]=b[top];
59 }
60 inline bool cmp(gap x,gap y)
61 {
62     return x.x<y.x;
63 }
64 int main()
65 {
66     fre("math");
67     int i,n=read();
68     for (i=1;i<=n;i++) a[i]=(struct gap){read(),i};
69     sort(a+1,a+i,cmp);
70     for (i=1;i<=n;i++)
71     {
72         b[a[i].num]=i;
73         p[i]=a[i].num;
74         a[i]=(struct gap){read(),i};
75     }
76     sort(a+1,a+i,cmp);
77     for (i=1;i<=n;i++) c[a[i].num]=p[i];
78     gb(1,n);
79     printf("%d\n",ans);
80     return 0;
81 }

 

 

  D1T3,题目:http://codevs.cn/problem/3287/

  题目要求路径上的所有边权的最小值最大

  对于两点之间的路径,不就是走最大的那几个边可以保证答案最大

  而在一棵树中是能保证图刚好连通的,做一颗最大生成树

  然后对于每组询问直接倍增求LCA即可,记得统计路径上的边权最小值

  1 // Copyright(c) Hadilo.All Rights Reserved.
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<cstring>
  5 #include<cmath>
  6 #include<iostream>
  7 #include<algorithm>
  8 #include<queue>
  9 #include<stack>
 10 #include<ctime>
 11 #define fre(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
 12 using namespace std;
 13 typedef long long LL;
 14 typedef double db;
 15 const long double CPS=CLOCKS_PER_SEC,TL=0.998;
 16 const int oo=2147483647,N=10001,M=50001,L=15,N2=20001;
 17 inline int read()
 18 {
 19     int re=0;
 20     bool fu=0;
 21     char ch=getchar();
 22     while ((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
 23     if (ch=='-')
 24     {
 25         fu=1;
 26         ch=getchar();
 27     }
 28     while (ch>='0'&&ch<='9')
 29     {
 30         re=re*10+ch-'0';
 31         ch=getchar();
 32     }
 33     return fu?-re:re;
 34 }
 35 struct gap
 36 {
 37     int u,v,w;
 38 };
 39 gap e[M];
 40 int first[N],f[N][L],g[N][L],next[N2],v[N2],w[N2],d[N],fa[N],t;
 41 inline int log2(int x)
 42 {
 43     int k=-1;
 44     while (x)
 45     {
 46         k++;
 47         x>>=1;
 48     }
 49     return k;
 50 }
 51 inline int find(int x)
 52 {
 53     return fa[x]==x?x:fa[x]=find(fa[x]);
 54 }
 55 inline void dfs(int x,int last)
 56 {
 57     if (last) fa[x]=find(last);
 58     t=max(t,d[x]=d[last]+1);
 59     f[x][0]=last;
 60     for (int i=first[x];i;i=next[i])
 61         if (v[i]!=last)
 62         {
 63             g[v[i]][0]=w[i];
 64             dfs(v[i],x);
 65         }
 66 }
 67 inline bool cmp(gap x,gap y)
 68 {
 69     return x.w>y.w;
 70 }
 71 int main()
 72 {
 73     fre("truck");
 74     int n=read(),m=read(),i,j,x,y,top=0,ans;
 75     for (i=1;i<=m;i++) e[i]=(struct gap){read(),read(),read()};
 76     sort(e+1,e+1+m,cmp);
 77     for (i=1;i<=n;i++) fa[i]=i;
 78     for (i=1;i<=m;i++)
 79     {
 80         x=find(e[i].u);
 81         y=find(e[i].v);
 82         if (x!=y)
 83         {
 84             fa[x]=y;
 85             top++;
 86             next[top]=first[v[top+n]=e[i].u];
 87             first[e[i].u]=top;
 88             next[top+n]=first[v[top]=e[i].v];
 89             first[e[i].v]=top+n;
 90             w[top]=w[top+n]=e[i].w;
 91         }
 92     }
 93     for (i=1;i<=n;i++) fa[i]=i;
 94     for (i=1;i<=n;i++) if (fa[i]==i) dfs(i,0);
 95     t=log2(t);
 96     for (i=1;i<=t;i++)
 97         for (j=1;j<=n;j++)
 98         {
 99             f[j][i]=f[f[j][i-1]][i-1];
100             g[j][i]=min(g[j][i-1],g[f[j][i-1]][i-1]);
101         }
102     m=read();
103     while (m--)
104     {
105         x=read();
106         y=read();
107         if (find(x)==find(y))
108         {
109             ans=oo;
110             if (d[x]<d[y]) swap(x,y);
111             for (i=log2(d[x]-d[y]);i>=0;i--)
112                 if (f[x][i]&&d[f[x][i]]>=d[y])
113                 {
114                     ans=min(ans,g[x][i]);
115                     x=f[x][i];
116                 }
117             if (x==y)
118             {
119                 printf("%d\n",ans);
120                 continue;
121             }
122             for (i=log2(d[x]);i>=0;i--)
123                 if (f[x][i]&&f[x][i]!=f[y][i])
124                 {
125                     ans=min(ans,min(g[x][i],g[y][i]));
126                     x=f[x][i];
127                     y=f[y][i];
128                 }
129             printf("%d\n",min(ans,min(g[x][0],g[y][0])));
130         }
131         else printf("-1\n");
132     }
133     return 0;
134 }

 

 

 

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posted @ 2016-11-08 15:57  Hadilo  阅读(273)  评论(0编辑  收藏