Educational Codeforces Round 14

A - Fashion in Berland

 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 = 201000;
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 a[1010];
38 int main(){
39     int n;
40     cin>>n;
41     int k=0;
42     for(int i=0;i<n;i++){
43          cin>>a[i];
44          if(!a[i]) k++;
45     }
46     if(n==1&&a[0]==0){
47         puts("NO");
48         exit(0);
49     }
50     if(n==1&&a[0]==1){
51         puts("YES");
52         exit(0);
53     }
54     if(k==1) cout<<"YES"<<endl;
55     else cout<<"NO"<<endl;
56     return 0;
57 }
View Code

 B - s-palindrome

题意:判断是否是 ‘镜像‘

 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 = 201000;
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 std::pair<char, char> mapP[] ={{'A', 'A'}, {'b', 'd'}, {'d', 'b'}, {'H', 'H'}, {'I', 'I'}, {'M', 'M'}, {'O', 'O'}, {'o', 'o'}, {'p', 'q'},
37      {'q', 'p'},{'T', 'T'}, {'U', 'U'}, {'V', 'V'}, {'v', 'v'},{'W', 'W'}, {'w', 'w'}, {'X', 'X'}, {'x', 'x'}, {'Y', 'Y'}
38  };
39 
40 bool check(char a, char b){
41     for(std::pair<char, char> p : mapP){
42        if(p.first == a && p.second == b){
43         return true;
44     }
45   }
46   return false;
47 }
48 
49 int main(){
50     string s;
51     cin>>s;
52     int i,j;
53     bool flag=false;
54     for(i=0,j=(int)s.length()-1;i<=j;i++,j--){
55          if(!check(s[i],s[j])){
56             flag=1;
57             puts("NIE");
58             break;
59          }
60     }
61     if(!flag)
62         puts("TAK");
63     return 0;
64 }
View Code

C - Exponential notation

题意:把一个值转换成 表达式a*10^b形式

字符处理

 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 = 201000;
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     string str="";
40     int ansb;
41     cin>>s;
42     int p=0;
43     while(p<(int)s.size()&&(s[p]=='0'||s[p]=='.')){
44         p++;
45     }
46     // cout<<p<<bndl;
47     str.push_back(s[p]);
48     str.push_back('.');
49     int pos=s.find('.');
50     if(pos==-1) pos=s.size();
51     for(int i=++p;i<s.size();i++){
52          if(s[i]!='.') str.push_back(s[i]);
53     }
54     if(pos==p) ansb=0;
55     else{
56          if(pos>p) ansb=pos-p;
57          else ansb=pos-p+1;
58     }
59     while(str.back()=='0'||str.back()=='.') str.pop_back();
60     if(ansb) str+="E"+to_string(ansb);
61     printf("%s\n",str.c_str());
62     return 0;
63 }
View Code

 D - Swaps in Permutation

题意:n个数(1,2。。。n) m种操作:交换下标为x y的数, 每种操作可选任意次,输出最大字典序。

思路:并查集。最后把每个位置上同一个块的从大到小输出

 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 = 1000010;
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 a[N];
37 int pre[N];
38 vector<int>ans[N];
39 vector<int>root;
40 int vis[N];
41 int pos[N];
42 int find(int x){
43      if(x==pre[x]) return x;
44      return pre[x]=find(pre[x]);
45 }
46 int main(){
47     std::ios::sync_with_stdio(false);
48     int n,m;
49     cin>>n>>m;
50     for(int i=1;i<=n;i++){
51          cin>>a[i];
52          pre[i]=i;
53     }
54     while(m--){
55          int x,y;
56          cin>>x>>y;
57          x=find(x),y=find(y);
58          if(x!=y) pre[x]=y;
59     }
60     for(int i=1;i<=n;i++){
61          int x=find(i);
62          ans[x].push_back(a[i]);
63          if(!vis[x]){
64              root.push_back(x);
65              vis[x]=true;
66          }
67     }
68     for(int i=0;i<(int)root.size();i++){ //从大到小排序
69         sort(ans[root[i]].rbegin(),ans[root[i]].rend());
70     }
71     for(int i=1;i<n;i++){
72          cout<<ans[find(i)][pos[find(i)]++]<<" ";
73     }
74     cout<<ans[find(n)][pos[find(n)]]<<endl;
75     return 0;
76 }
View Code

 E - Xor-sequences

题意:n个数,选k个出来,形成一个新的序列,该序列相邻两个数异或后数的二进制中1的个数是3的倍数。问有几个序列

思路:每个数转化成一个点,两个数满足 异或后数的二进制1的个数是3倍数的则两点之间建一条边。转化成求边的条数。

矩阵快速幂

M(i,j):(ai^aj)%3==0 则最后答案是M^(k-1)的元素和

借鉴了Q神的模板

 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 = 1000010;
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 
38 struct Matrix{
39     LL a[110][110];
40     Matrix(){
41         clc(a,0);
42     } 
43     void init(){
44         for(int i=0;i<100;i++) a[i][i]=1;
45     }
46     Matrix operator * (const Matrix &B)const{
47          Matrix C;
48          for(int i=0;i<100;i++){
49              for(int k=0;k<100;k++){
50                  for(int j=0;j<100;j++){
51                      (C.a[i][j]+=a[i][k]*B.a[k][j])%=MOD;
52                  }
53              }
54          }
55          return C;
56     }
57     Matrix operator ^ (const LL &t) const{
58         Matrix A= (*this),res;
59         res.init();
60         LL p=t;
61         while(p){
62             if(p&1) res=res*A;
63             A=A*A;
64             p>>=1;
65         }
66         return res;
67     }
68 };
69 LL a[110];
70 int main(){
71     LL n,k;
72     scanf("%I64d%I64d",&n,&k);
73     for(int i=0;i<n;i++){
74          scanf("%I64d",&a[i]);
75     }
76     Matrix A;
77     for(int i=0;i<n;i++){
78         for(int j=0;j<n;j++){
79              if(__builtin_popcountll(a[i]^a[j])%3==0)
80                 A.a[i][j]=1;
81         }
82     }
83     LL ans=0;
84     A=A^(k-1);
85     for(int i=0;i<n;i++){
86         for(int j=0;j<n;j++){ 
87             (ans+=A.a[i][j])%=MOD;
88         }
89     }
90     printf("%I64d\n",ans);
91     return 0;   
92 }
View Code

 F - Couple Cover

题意:n个数,任选两个相乘,q次询问,每次输出不小于x的乘积对数

思路:总的方案数-小于x

预处理每种面积的方案数

 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 
37 int cnt[N];
38 int maxn=3*1e6;
39 LL a[N];
40 LL sum[N];
41 void init(){ 
42     for(int i=1;i<=maxn;i++){ 
43         for(int j=1;j*i<=maxn;j++){ 
44            if(i==j) a[i*j]+=(LL)cnt[i]*(cnt[i]-1);
45            else a[i*j]+=(LL)cnt[i]*cnt[j];
46         } 
47     }
48     for(int i=1;i<=maxn;i++){ 
49         sum[i]=sum[i-1]+a[i];
50     }
51 }
52 int main(){
53     int n;
54     scanf("%d",&n);
55     for(int i=0;i<n;i++){
56         int x;
57         scanf("%d",&x);
58         if(x<=maxn)
59         cnt[x]++;
60     }
61     init();
62     int q;
63     scanf("%d",&q);
64     while(q--){
65          int x;
66          scanf("%d",&x);
67          LL ans=(LL)n*(n-1)-sum[x-1];
68          printf("%I64d\n",ans);
69     }
70     return 0;
71 }
View Code

 

posted @ 2016-07-14 21:02  yyblues  阅读(251)  评论(0编辑  收藏  举报