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XDOJ_1006_^&|

http://acm.xidian.edu.cn/problem.php?id=1006

 

这道题真的很厉害,开始一看,数据这么大,写了线段树和dp,可想而知都超时了。

要询问这么多次,于是向三个运算符号研究,发现了一种把O(n)处理完的方法,把数都分成二进制位,a[i]保存所有以i结尾的结果为1的个数和,b[i],c[i]类似。

还有一点,a[i]*对应的进制时可能会超int,被这个也坑了好久。

 

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int bin[100005][35],two[35],n;double a[100005],b[100005],c[100005];int main(){    two[0] = 1;    for(int i = 1;i <= 30;i++)  two[i] = two[i-1]*2;    while(~scanf("%d",&n))    {        memset(bin,0,sizeof(bin));        int maxcnt = 0;        double ans1 = 0,ans2 = 0,ans3 = 0;        for(int i = 1;i <= n;i++)        {            int x;            scanf("%d",&x);            ans1 -= x;            ans2 -= x;            ans3 -= x;            int cnt = 0;            while(x)            {                bin[i][++cnt] = x%2;                x /= 2;            }            maxcnt = max(cnt,maxcnt);        }        for(int j = 1;j <= maxcnt;j++)        {            int last0 = 0,last1 = 0;            for(int i = 1;i <= n;i++)            {                if(bin[i][j] == 1)                {                    if(last1 > 0)   a[i] = a[last1-1]+i-last1;                    else            a[i] = i;                    b[i] = i-last0;                    c[i] = i;                    last1 = i;                }                else                {                    if(i > 1)   a[i] = a[i-1];                    else        a[i] = 0;                    b[i] = 0;                    c[i] = last1;                    last0 = i;                }                ans1 += a[i]*two[j];                ans2 += b[i]*two[j];                ans3 += c[i]*two[j];            }        }        printf("%.3lf %.3lf %.3lf\n",ans1/n/n,ans2/n/n,ans3/n/n);    }    return 0;}

 

XDOJ_1006_^&|