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[暴力 乱搞] 51Nod 1674 算法马拉松19 A 区间的价值 V2

2016-10-31 18:08 405 查看
因为每个点向前and 向前or 的后缀值是O(32) 

所以可以暴力分段求和 32n

类似的还有后缀gcd 是O(logV)的

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()) x*=b;
}

const ll P=1000000007;
const int N=100005;

struct abcd{
int l,r,val;
abcd(int l=0,int r=0,int val=0):l(l),r(r),val(val) { }
}And[35],Or[35];
int tota,toto;

int n,a
;
ll ans;

int main(){
int pnt=0;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); for (int i=1;i<=n;i++) read(a[i]);
for (int i=1;i<=n;i++){

for (int j=1;j<=tota;j++) And[j].val&=a[i];
And[++tota]=abcd(i,i,a[i]);
pnt=0;
for (int j=1;j<=tota;j++){
int l=And[j].l;
while (j+1<=tota && And[j+1].val==And[j].val) j++;
And[++pnt]=abcd(l,And[j].r,And[j].val);
}
tota=pnt;

for (int j=1;j<=toto;j++) Or[j].val|=a[i];
Or[++toto]=abcd(i,i,a[i]);
pnt=0;
for (int j=1;j<=toto;j++){
int l=Or[j].l;
while (j+1<=toto && Or[j+1].val==Or[j].val) j++;
Or[++pnt]=abcd(l,Or[j].r,Or[j].val);
}
toto=pnt;

int pnt1=1,pnt2=1;
for (int j=1;j<=i;){
while (j>And[pnt1].r) pnt1++;
while (j>Or[pnt2].r) pnt2++;
int last=min(And[pnt1].r,Or[pnt2].r);
ans+=(ll)And[pnt1].val*Or[pnt2].val%P*(last-j+1)%P;
ans%=P;
j=last+1;
}
}
printf("%lld\n",ans);
return 0;
}
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