51nod1674 区间的价值

题意:输入一个序列,区间的价值定义为这个区间的所有数&乘以所有数|,问所有区间的价值为多少?

题解:这里要知道,f1(1, j) = and(a[1], a[2]... a[j])随着j增大f(i, j)的1个数会减少或不变,所以最后f的值最多只有log(n)个

分治算答案.对于一个区间,左右分治,再计算跨越mid的区间的价值,右边的值最多有log(len)个那么可以分块计算答案,复杂度n*logn*logn

#include <bits/stdc++.h>
#define maxn 101000
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const ll mod = 1e9+7;
int aor[maxn], aand[maxn], num[maxn], a[maxn];
ll f(int l,int r){
    if(l == r) return (ll)a[l]*a[l];
    int mid = (l+r)>>1;
    ll ans = 0;
    ans = (ans+f(l, mid)+f(mid+1, r))%mod;
    int n = 1;
    num[n] = 1;
    aor[n] = aand[n] = a[mid+1];
    for(int i=mid+2;i<=r;i++){
        if(((aor[n]|a[i]) != aor[n])||((aand[n]&a[i]) != aand[n])){
            n++;
            aor[n] = aor[n-1]|a[i];
            aand[n] = aand[n-1]&a[i];
            num[n] = 1;
        }
        else num[n]++;
    }
    int t1 = a[mid], t2 = a[mid];
    for(int i=mid;i>=l;i--){
        t1 &= a[i], t2 |= a[i];
        for(int j=1;j<=n;j++)
            ans = (ans+(ll)(t1&aand[j])*num[j]%mod*(t2|aor[j])%mod)%mod;
    }
    return (ans+mod)%mod;
}
int main(){
    int n;
    scanf("%d", &n);
    for(ll i=1;i<=n;i++)
        scanf("%d", &a[i]);
    printf("%lld\n", f(1, n));
    return 0;
}