• AcWing周赛 6


    A:水题。

    https://www.acwing.com/problem/content/3736/

     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 using namespace std;
     5 const int N=110;
     6 int w[N];
     7 int main()
     8 {
     9     int n;
    10     cin>>n;
    11     int sum=0;
    12     for(int i=0;i<n;i++) cin>>w[i],sum+=w[i];
    13     int res=0;
    14     if(sum%2==0){
    15         for(int i=0;i<n;i++)
    16             if(w[i]%2==0)
    17                 res++;
    18     }else{
    19         for(int i=0;i<n;i++)
    20             if(w[i]%2==1)
    21                 res++;
    22     }
    23     cout<<res;
    24     return 0;
    25 }

    B:https://www.acwing.com/problem/content/3737/

    f ( i )表示大于等于 i 的且只包含4和7的最小的数,求f(l)+f(l+1)+...+f(r)的值。

    首先枚举出全部的可能的数,可以发现1~1e9只会有2+2^2+...+2^10个数,即最多2048个数。

    而且最终的答案是可以分区间来计算的,故只会计算2^11次。

    考试的时候的代码:

     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 using namespace std;
     5 typedef long long LL;
     6 LL a[12];
     7 LL cal(){//计算数组a对应的值
     8     LL res=0;
     9     for(LL i=10;i>=0;i--){
    10         if(a[i]==1)
    11             res=res*10+4;
    12         else if(a[i]==2)
    13             res=res*10+7;
    14         else
    15             res=res*10;
    16     }
    17 }
    18 int main()
    19 {
    20     LL l,r;
    21     cin>>l>>r;
    22     a[0]=1;
    23     LL res=0;
    24     for(LL i=l;i<=r;){//i表示现在的左端点
    25         LL t=cal();
    26         if(t>=r) res+=(r-i+1)*t,i=r+1;
    27         else if(t>=i) res+=(t-i+1)*t,i=t+1;
    28         a[0]++;
    29         for(LL j=0;j<=10;j++)//不好用二进制,所以用三进制,用1表示4,2表示7
    30             if(a[j]==3)
    31             {
    32                 a[j]=1;
    33                 a[j+1]+=1;
    34             }
    35     }
    36     cout<<res;
    37     return 0;
    38 }

    更简洁的代码(取自yxc):

     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 #include <vector>
     5 using namespace std;
     6 typedef long long LL;
     7 vector<LL> S;
     8 void dfs(LL u,LL n){
     9     S.push_back(n);
    10     if(u==10) return ;
    11     dfs(u+1,n*10+4);
    12     dfs(u+1,n*10+7);
    13 }
    14 int main(void){
    15     dfs(0,0);
    16     sort(S.begin(),S.end());
    17     LL l,r;
    18     cin>>l>>r;
    19     LL res=0;
    20     for(int i=1;i<=S.size();i++){
    21         LL a=S[i-1]+1,b=S[i];
    22         S[i]*max(0LL,min(r,b)-max(l,a));
    23         res+=S[i]*max(0LL,min(r,b)-max(l,a)+1);
    24     }
    25     cout<<res;
    26     return 0;
    27 }

    C:状态压缩DP,太复杂了,不会。

    https://www.acwing.com/problem/content/3738/

    抄自yxc:

     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 
     5 #define x first
     6 #define y second
     7 
     8 using namespace std;
     9 
    10 typedef pair<int, int> PII;
    11 
    12 const int N = 22, M = 1 << N;
    13 
    14 int n, m;
    15 int e[N];
    16 int f[M];
    17 PII g[M];
    18 
    19 int main()
    20 {
    21     cin >> n >> m;
    22 
    23     if (m == n * (n - 1) / 2)
    24     {
    25         puts("0
    ");
    26         return 0;
    27     }
    28 
    29     for (int i = 0; i < n; i ++ ) e[i] = 1 << i;
    30 
    31     while (m -- )
    32     {
    33         int a, b;
    34         cin >> a >> b;
    35         a --, b -- ;
    36         e[a] |= 1 << b;
    37         e[b] |= 1 << a;
    38     }
    39 
    40     memset(f, 0x3f, sizeof f);
    41     for (int i = 0; i < n; i ++ ) f[e[i]] = 1, g[e[i]] = {0, i};
    42     for (int i = 0; i < 1 << n; i ++ )
    43     {
    44         if (f[i] == 0x3f3f3f3f) continue;
    45         for (int j = 0; j < n; j ++ )
    46         {
    47             if (i >> j & 1)
    48             {
    49                 int k = i | e[j];
    50                 if (f[k] > f[i] + 1)
    51                 {
    52                     f[k] = f[i] + 1;
    53                     g[k] = {i, j};
    54                 }
    55             }
    56         }
    57     }
    58 
    59     int k = (1 << n) - 1;
    60     cout << f[k] << endl;
    61     while (k)
    62     {
    63         cout << g[k].y + 1 << ' ';
    64         k = g[k].x;
    65     }
    66     return 0;
    67 }
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  • 原文地址:https://www.cnblogs.com/greenofyu/p/14979408.html
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