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首先将数列按位分解成32个位数列.

枚举区间的左端点,,区间的and值要对答案有贡献必须为1,随着右端点右移,and值为不递增的,而区间的or值要对答案有贡献必须为1,随着右端点右移,or值为不递减的。

可以求出每一段一段的最大长度然后统计

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100011
#define mod 1000000007
int n,p[66],L[33][N],R[33][N];
bool a[33][N];
ll ans=0;
int main(){
    int x;
    in>>n;
    Fur(i,1,n){
        in>>x;
        Fur(j,0,32)
            a[j][i]=x&1,x>>=1;
    }

    p[0]=1;Fur(i,1,64)p[i]=(p[i-1]<<1)%mod;

    Fur(i,0,32){
        int l=1,r=1;
        while(l<=n){
            if(!a[i][l])L[i][l]=0;
            else{
                r=MAX(r,l);
                while(r<n&&a[i][r+1])++r;
                L[i][l]=r;
            }
            ++l;
        }
    }

    Fur(i,0,32){
        int l=1,r=1;
        while(l<=n){
            if(a[i][l])R[i][l]=l;
            else{
                r=MAX(r,l);
                while(r<=n&&!a[i][r])++r;
                R[i][l]=r;
            }
            ++l;
        }
    }

    Fur(i,0,32)
        Fur(j,1,n)
        if(a[i][j])
            Fur(k,0,32)
                if(R[k][j]<=L[i][j])
                ans=(ans+(ll)(L[i][j]-R[k][j]+1)*p[i+k])%mod;

    cout<<ans<<endl;
}
51nod 1674 区间的价值 V2
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