bzoj4750

单调栈+前缀和

max很明显用单调栈搞，但是异或和呢？异或和我们拆位，对于每段区间的异或和[l[i]-i],[i,r[i]]答案就是0->1,1->0的乘积，但是统计的时候事实上是[l[i]-2,i-1],因为异或和本身是前缀和，所以要-1，单调栈又是一个前缀和，也要-1，所以就是-2

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + , mod = 1e9 + ;
int n;
int l[N], r[N], st[N];
ll a[N], sum[N][];
void up(ll &x, const ll &t)
{
    x = ((x + t) % mod + mod) % mod;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int top = ;
        ll ans = , tot = ;
        scanf("%d", &n);
        for(int i = ; i <= n; ++i)
        {
            scanf("%lld", &a[i]);
            tot = tot ^ a[i];
            for(int j = ; j < ; ++j) sum[i][j] = sum[i - ][j] + ((tot & ( << j)) > );
        }
        for(int i = ; i <= n; ++i)
        {
            l[i] = r[i] = i;
            while(top && a[i] > a[st[top]])
            {
                r[st[top - ]] = r[st[top]];
                l[i] = l[st[top]];
                --top;
            }
            st[++top] = i;
        }
        while(top) r[st[top - ]] = r[st[top]], --top;
        for(int i = ; i <= n; ++i)
        {
            ll pw = ;
            for(int j = ; j < ; ++j)
            {
                ll tmp1 = sum[r[i]][j] - sum[i - ][j], tmp2 = sum[i - ][j] - sum[max(l[i] - , )][j];
                up(ans, tmp1 * (ll)(i - l[i] +  - tmp2) % mod * pw % mod * a[i] % mod);
                up(ans, (ll)(r[i] - i +  - tmp1) * tmp2 % mod * pw % mod * a[i] % mod);
                pw = pw *  % mod;
            }
        }
        printf("%lld\n", ans);
    }
    return ;
}