2017.11.1模拟 signal-CFANZ编程社区

http://www.elijahqi.win/archives/1460

#include<cmath> 
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 110000
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=gc();}
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x*f;
}
long long ans1,ans2,ans3,sum1;int n,a[N],bin[35],sum[N],max1,s[2];
inline int calc_even(int x){return x/2+1;}
inline int calc_odd(int x){return x/2;}
int main(){
//  freopen("signal.in","r",stdin);
    n=read();
    for (int i=1;i<=n;++i) a[i]=read(),max1=max(max1,a[i]),sum1+=a[i];
    for (int i=0;i<=31;++i) bin[i]=1<<i;int Log=log2(max1);
    for (int i=0;i<=Log;++i){
        int pos3=1;long long count3=0;
        for (int j=1;j<=n;++j){
            if (j>pos3) pos3++;
            while ((bin[i]&a[pos3])==0&&pos3<=n) ++pos3;
            count3+=n-pos3+1;
        }ans3+=(long long)bin[i]*count3;
        int pos2=1;long long count2=0;
        for (int j=1;j<=n;++j){
            if (j>pos2) pos2++;
            while((bin[i]&a[pos2])&&pos2<=n) ++pos2;
            if (a[j]&bin[i]) count2+=pos2-j;
        }ans2+=(long long)bin[i]*count2;memset(sum,0,sizeof(sum));long long count1=0;
        for (int j=n;j>=1;--j) sum[j]+=sum[j+1]+(a[j]&bin[i]);int pos1=1;
        while ((a[pos1]&bin[i])==0&&pos1<=n)++pos1;int tmp=0;
        while(1){
            if (pos1>n) break;pos1++;tmp++;int last=pos1;
            while((a[pos1]&bin[i])==0&&pos1<=n)++pos1;
            s[tmp%2]+=pos1-last+1;
        }pos1=1;
        while ((a[pos1]&bin[i])==0&&pos1<=n)++pos1;int last=pos1;tmp=0;
        for (int j=1;j<=n;++j){
            if (a[j]&bin[i]) {
                ++tmp;last=j;pos1=j+1;
                while((a[pos1]&bin[i])==0&&pos1<=n)++pos1;
                count1+=s[tmp%2];s[tmp%2]-=pos1-last;
            }else{
                count1+=s[(tmp+1)%2];
            }
        }ans1+=(long long)bin[i]*count1;
    }
    ans1<<=1;ans1-=sum1;ans2<<=1;ans2-=sum1;ans3<<=1;ans3-=sum1;
    printf("%.3f %.3f %.3f",(double)ans1/((long long)n*n),(double)ans2/((long long)n*n),(double)ans3/((long long)n*n));
    return 0;
}