bzoj4750

Posted 2020-10-12 123456

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 + 5, mod = 1e9 + 61;
int n;
int l[N], r[N], st[N];
ll a[N], sum[N][31]; 
void up(ll &x, const ll &t) 
{
    x = ((x + t) % mod + mod) % mod;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int top = 0;
        ll ans = 0, tot = 0;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) 
        {
            scanf("%lld", &a[i]);
            tot = tot ^ a[i];
            for(int j = 0; j < 31; ++j) sum[i][j] = sum[i - 1][j] + ((tot & (1 << j)) > 0);
        }
        for(int i = 1; i <= n; ++i)
        {
            l[i] = r[i] = i;
            while(top && a[i] > a[st[top]]) 
            {
                r[st[top - 1]] = r[st[top]];
                l[i] = l[st[top]];
                --top;
            }
            st[++top] = i;
        }
        while(top) r[st[top - 1]] = r[st[top]], --top;      
        for(int i = 1; i <= n; ++i)
        {
            ll pw = 1;
            for(int j = 0; j < 30; ++j)
            {
                ll tmp1 = sum[r[i]][j] - sum[i - 1][j], tmp2 = sum[i - 1][j] - sum[max(l[i] - 2, 0)][j];
                up(ans, tmp1 * (ll)(i - l[i] + 1 - tmp2) % mod * pw % mod * a[i] % mod);
                up(ans, (ll)(r[i] - i + 1 - tmp1) * tmp2 % mod * pw % mod * a[i] % mod);
                pw = pw * 2 % mod;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}

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