2021“MINIEYE杯”中国大学生算法设计超级联赛（2）1011.I love max and multiply (思维,位运算)

• 题意:有两个长度为\(n\)的序列\(a\)和\(b\),定义\(C_k=max(A_i,B_j)\ (i\ and\ j\ge k)\),求\(\sum^{n-1}_{i=0}C_i\).

• 题解:暴力思路:求出所有的\(C_k\),然后从\(n-1\)倒着维护最大值贡献给答案即可.

  根据到这维护最大值这个思想,我们考虑\(i\)&\(j\)=\(i\)的情况,根据&运算的性质,两个数&之后只会不变或者变小,我们倒着枚举\(i\),根据上述性质,我们找\([i,n-1]\)中满足条件的\(j\),这里我们只要看单独的某一个\(1\)即可,可以用dp转移过来,那么我们每次可以得到\([i,n-1]\)中的满足条件的\(j\),更新最大值最小值.最后在求答案贡献的时候,\(A[i]\)表示的是\([i,n-1]\)中的\(i\)&\(j\)=\(i\)的某个位置\(x_1\)的值,同理,\(B[i]\)表示的是某个位置\(x_2\)的值,因为我们在更新的时候保证\(i\)&\(j\)=\(i\),且根据性质\(j\)一定不小于\(i\),那么\(x_1\)&\(x_2\ge i\),然后就根据我们前面的暴力思路求贡献给答案就好了.

• 代码:

  #include <iostream>
  #include <iomanip>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define db double
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 998244353;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  int gcd(int a,int b){return b?gcd(b,a%b):a;}
  int lcm(int a,int b){return a/gcd(a,b)*b;}
  
  ll a[N],A[N],b[N],B[N];
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      #ifdef lr001
          freopen("/Users/somnus/Desktop/data/in.txt","r",stdin);
      #endif
      #ifdef lr002
          freopen("/Users/somnus/Desktop/data/out.txt","w",stdout);
      #endif
      int _;
      cin>>_;
      while(_--){
          int n;
          cin>>n;
          rep(i,0,n-1) cin>>a[i],A[i]=a[i];
          rep(i,0,n-1) cin>>b[i],B[i]=b[i];
          int base=0;
          while((1<<base)<=n) base++;  //要包括n
          /* brute_force
          rep(i,0,n-1){
              rep(j,0,n-1){
                  res[i&j]=max(res[i&j],a[i]*b[j]);
              }
          }
          int ans=0;
          per(i,n-1,0){
              res[i]=max(res[i],res[i+1]);
              ans+=res[i];
          }
          cout<<ans<<'\n';
          */
          rep(i,n,(1<<base)-1){
              A[i]=B[i]=-INF;
              a[i]=b[i]=INF;
          }
          per(i,n,0){
              rep(j,0,base-1){
                  int cur=1<<j;
                  if((i&cur)==0){
                      A[i]=max(A[i],A[i^cur]);
                      B[i]=max(B[i],B[i^cur]);
                      a[i]=min(a[i],a[i^cur]);
                      b[i]=min(b[i],b[i^cur]);
                  }
              }
          }
          ll mx=-1e18;
          ll ans=0;
          per(i,n-1,0){
              ll res=-1e18;
              res=max(res,A[i]*B[i]);
              res=max(res,A[i]*b[i]);
              res=max(res,a[i]*B[i]);
              res=max(res,a[i]*b[i]);
              res=max(res,mx);
              mx=res;
              ans=(ans+res+mod)%mod;
          }
          cout<<ans<<'\n';
      }
  
  
  
      return 0;
  }
  
  

标签：love,gcd,int,max,cin,multiply,const,include,define
来源： https://www.cnblogs.com/lr599909928/p/15080639.html