Codeforces Round #362

 1 #pragma comment(linker, "/STACK:102c000000,102c000000")
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <sstream>
 6 #include <string>
 7 #include <algorithm>
 8 #include <list>
 9 #include <map>
10 #include <vector>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <cstdlib>
15 // #include <conio.h>
16 using namespace std;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 #define inf 0x3f3f3f3f
19 #define lson l,mid,rt<<1
20 #define rson mid+1,r,rt<<1|1
21 const int N = 3*1e6+10;
22 const int MOD = 1e9+7;
23 #define LL long long
24 #define mi() (l+r)>>1
25 double const pi = acos(-1);
26 
27 void fre() {
28     freopen("in.txt","r",stdin);
29 }
30 
31 // inline int r() {
32 //     int x=0,f=1;char ch=getchar();
33 //     while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
34 //     while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;
35 // }
36 int main(){
37     int t,s,x;
38     cin>>t>>s>>x;
39     bool flag=false;
40     if(x<t) puts("NO"),exit(0);
41     if(((x-t)%s==0)||((x-t-1)%s==0&&x-t-1!=0)) flag=true;
42     if(flag) puts("YES");
43     else puts("NO");
44     return 0;
45 }

View Code

B - Barnicle

和上场有道差不多

 1 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <sstream>
 6 #include <string>
 7 #include <algorithm>
 8 #include <list>
 9 #include <map>
10 #include <vector>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <cstdlib>
15 // #include <conio.h>
16 using namespace std;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 #define inf 0x3f3f3f3f
19 #define lson l,mid,rt<<1
20 #define rson mid+1,r,rt<<1|1
21 const int N = 1e7+10;
22 const int MOD = 1e9+7;
23 #define LL long long
24 #define mi() (l+r)>>1
25 double const pi = acos(-1);
26 
27 void fre() {
28     freopen("in.txt","r",stdin);
29 }
30 
31 // inline int r() {
32 //     int x=0,f=1;char ch=getchar();
33 //     while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
34 //     while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;
35 // }
36 
37 int main(){
38     string s;
39     int l=0;
40     cin>>s;
41     int pos=s.find('e');
42     // cout<<pos<<endl;
43     for(int j=pos+1;j<=s.size()-1;j++){
44             l=s[j]-'0'+l*10;
45     }
46     // cout<<l<<endl;
47     string str="";
48     int pos2=s.find('.');
49     // cout<<pos2<<endl;
50     for(int i=0;i<pos2;i++){
51          str+=s[i];
52          // cout<<str<<endl; 
53          if(str[0]=='0'||l==0){
54          str+='.';
55          }
56     }
57     for(int i=pos2+1;i<pos;i++){
58          str+=s[i];
59          if(i-pos2==l)
60             str+='.';
61     }
62     if(l>pos-pos2-1){
63        for(int i=0;i<l-pos+pos2+1;i++){
64            str+='0';
65        }
66     }
67     if(str.find('.')!=-1){
68         while(str.back()=='.'||str.back()=='0') str.pop_back();
69     }
70     if(str=="")
71     str='0';
72     if(l==0){
73         cout<<str<<endl;
74         exit(0);
75     }
76     int p=0;
77     for(int i=0;i<(int)str.size();i++){
78         if(str[i]!='0'&&str[i]!='.'){
79            p=i;
80            break;
81         }
82     }
83     for(int i=p;i<(int)str.size();i++){
84          cout<<str[i];
85     }
86     cout<<endl;
87     return 0;
88 }

View Code

C - Lorenzo Von Matterhorn

题意：完全二叉树 q次询问，op=1时，在x到y的最短边上c，op=2，输出x到y的权值和

思路：应为是完全二叉树，直接暴力每条边，加上权值，用mp存当前点到父亲的权值

 1 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <sstream>
 6 #include <string>
 7 #include <algorithm>
 8 #include <list>
 9 #include <map>
10 #include <vector>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <cstdlib>
15 // #include <conio.h>
16 using namespace std;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 #define inf 0x3f3f3f3f
19 #define lson l,mid,rt<<1
20 #define rson mid+1,r,rt<<1|1
21 const int N = 1e7+10;
22 const int MOD = 1e9+7;
23 #define LL long long
24 #define mi() (l+r)>>1
25 double const pi = acos(-1);
26 
27 void fre() {
28     freopen("in.txt","r",stdin);
29 }
30 
31 // inline int r() {
32 //     int x=0,f=1;char ch=getchar();
33 //     while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
34 //     while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;
35 // }
36 map<LL,LL>mp;
37 void add(LL x,LL y,LL c){
38      while(x!=y){
39         if(x>y) swap(x,y);
40           mp[y]+=c;
41           y>>=1;
42      }
43 }
44 
45 LL Sum(LL x,LL y){
46      LL sum=0;
47      while(x!=y){
48          if(x>y) swap(x,y);
49          sum+=mp[y];
50          y>>=1;
51      }
52      return sum;
53 }
54 int main(){
55     // fre();
56     int T;
57     scanf("%d",&T);
58     mp.clear();
59     while(T--){
60        int op;
61        LL x,y,c;
62        scanf("%d",&op);
63        if(op==1){
64            scanf("%I64d%I64d%I64d",&x,&y,&c);
65            add(x,y,c);
66        }
67        else{
68            scanf("%I64d%I64d",&x,&y);
69            LL ans=Sum(x,y);
70            printf("%I64d\n",ans);
71        }
72     }
73     return 0;
74 }

View Code

D - Puzzles

题意：n个节点，n-1个数，pi是i+1的父亲。问每个点的在随机DFS中访问次数的期望

题意：理解什么是随机dfs访问期望就简单了：非当前点的父亲节点和子孙节点的点，出现在该点的前后概率均为0.5；

所以dp[i]=dp[f]+1+0.5*(nodes[f]-nodes[i]-1);

f:i的父亲

nodes[]:i为根的树所含节点总数

所以，先处理每个点的子孙节点总数，再dp

对于当前点来说，父亲一定在dfs之前出现，所以次数+1,子孙不算，其余点要么在前要么在后，因此次数+0.5

 1 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <sstream>
 6 #include <string>
 7 #include <algorithm>
 8 #include <list>
 9 #include <map>
10 #include <vector>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <cstdlib>
15 // #include <conio.h>
16 using namespace std;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 #define inf 0x3f3f3f3f
19 #define lson l,mid,rt<<1
20 #define rson mid+1,r,rt<<1|1
21 const int N = 1e5+10;
22 const int MOD = 1e9+7;
23 #define LL long long
24 #define mi() (l+r)>>1
25 double const pi = acos(-1);
26 
27 void fre() {
28     freopen("in.txt","r",stdin);
29 }
30 
31 // inline int r() {
32 //     int x=0,f=1;char ch=getchar();
33 //     while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
34 //     while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;
35 // }
36 vector<int>g[N];
37 int nodes[N];
38 double dp[N];
39 int pre[N];
40 int dfs(int f){
41      if(g[f].size()==0){
42         return nodes[f]=1;
43      }
44      for(int i=0;i<g[f].size();i++){
45          int v=g[f][i];
46          dfs(v);
47          nodes[f]+=nodes[v];
48     }
49     nodes[f]++;
50 }
51 
52 double DP(int f){
53     if(g[f].size()==0){
54         return 0;
55     }
56     for(int i=0;i<g[f].size();i++){
57         int v=g[f][i];
58         dp[v]=dp[f]+1+0.5*(nodes[f]-nodes[v]-1);
59         DP(v);
60     }
61 }
62 int main(){
63     fre();
64     int n;
65     scanf("%d",&n);
66     pre[1]=-1;
67     for(int i=1;i<=n+1;i++) g[i].clear();
68     for(int i=2;i<=n;i++){
69          int x;
70          scanf("%d",&x);
71          if(x==i) continue;
72          g[x].push_back(i);
73          pre[i]=x;
74     }
75     dfs(1);
76     dp[1]=1.0;
77     DP(1);
78     for(int i=1;i<n;i++){
79          printf("%.1lf ",dp[i]);
80     }
81     printf("%.1lf\n",dp[n]);
82     return 0;
83 }

View Code

posted @ 2016-07-16 10:41 yyblues 阅读(351) 评论(0) 编辑收藏举报

刷新页面返回顶部

blues

Ever tried,ever failed,no matter,try again,fail again,fail better.

Codeforces Round #362

公告