hdu5751 Eades

今天热身考到FFT，完全忘光了，模板敲错了。。。
晚上温习下以前的题目

这题就是从最大值每次分割现在的区间，这样递归的区间最大值会更小，对于每种最大值都是卷积做

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 60005;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int n;
int a[MAXN];
ll cnt[MAXN];
vector<int> po[MAXN]; // x的所有位置

/*********Segtree************/
int tree[MAXN<<2];
void Build(int l,int r,int rt){
    if(l == r) {
        tree[rt] = a[l]; return;
    }
    int m = (l+r)>>1;
    Build(lson); Build(rson);
    tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);
}
int Query(int L,int R,int l,int r,int rt){
    if(L <= l && r <= R) return tree[rt];
    int m = (l+r)>>1;
    int ans = -1;
    if(L <= m) ans = max(ans, Query(L,R,lson) );
    if(R > m) ans = max(ans, Query(L,R,rson) );
    return ans;
}
/***************FFT**********/
int A[MAXN<<2], B[MAXN<<2]; int C[MAXN<<2];
namespace FFT {
    int pos[MAXN<<2];
    struct comp {
        double r , i ;
        comp ( double _r = 0 , double _i = 0 ) : r ( _r ) , i ( _i ) {}
        comp operator + ( const comp& x ) {
            return comp ( r + x.r , i + x.i ) ;
        }
        comp operator - ( const comp& x ) {
            return comp ( r - x.r , i - x.i ) ;
        }
        comp operator * ( const comp& x ) {
            return comp ( r * x.r - i * x.i , i * x.r + r * x.i ) ;
        }
        comp conj () {
            return comp ( r , -i ) ;
        }
    } A[MAXN<<2] , B[MAXN<<2] ;

    const double pi = acos ( -1.0 ) ;
    void FFT ( comp a[] , int n , int t ) {
        for ( int i = 1 ; i < n ; ++ i ) if ( pos[i] > i ) swap ( a[i] , a[pos[i]] ) ;
        for ( int d = 0 ; ( 1 << d ) < n ; ++ d ) {
            int m = 1 << d , m2 = m << 1 ;
            double o = pi * 2 / m2 * t ;
            comp _w ( cos ( o ) , sin ( o ) ) ;
            for ( int i = 0 ; i < n ; i += m2 ) {
                comp w ( 1 , 0 ) ;
                for ( int j = 0 ; j < m ; ++ j ) {
                    comp& A = a[i + j + m] , &B = a[i + j] , t = w * A ;
                    A = B - t ;
                    B = B + t ;
                    w = w * _w ;
                }
            }
        }
        if ( t == -1 ) for ( int i = 0 ; i < n ; ++ i ) a[i].r /= n ;
    }
    void mul ( int *a , int *b , int *c ,int k) {
        int i , j ;
        for ( i = 0 ; i < k ; ++ i ) A[i] = comp ( a[i] , b[i] ) ;
        j = __builtin_ctz ( k ) - 1 ;
        for ( int i = 0 ; i < k ; ++ i ) {
            pos[i] = pos[i >> 1] >> 1 | ( ( i & 1 ) << j ) ;
        }
        FFT ( A , k , 1 ) ;
        for ( int i = 0 ; i < k ; ++ i ) {
            j = ( k - i ) & ( k - 1 ) ;
            B[i] = ( A[i] * A[i] - ( A[j] * A[j] ).conj () ) * comp ( 0 , -0.25 ) ;
        }
        FFT ( B , k , -1 ) ;
        for ( int i = 0 ; i < k ; ++ i ) {
            c[i] = ( long long ) ( B[i].r + 0.5 ) ;
        }
    }
}
/**************cdq***********/
void cdq(int l,int r){
    if(l > r) return;
    if(l == r) { cnt[1]++; return; }
    int num = Query(l,r,1,n,1);
    int st = lower_bound(po[num].begin(), po[num].end(), l) - po[num].begin();
    int ed = upper_bound(po[num].begin(), po[num].end(), r) - po[num].begin()-1;
    int m = 0;
    for(int i = st; i <= ed+1; ++i){
        int le = st==i? l-1 : po[num][i-1];
        int re = ed+1==i? r+1 : po[num][i];
        A[m++] = re-le;
    }
    int len = 1;
    while(len < 2*m) len<<=1;
    for(int i = 0; i < m; ++i) B[i] = A[m-1-i];
    for(int i = m; i < len; ++i) A[i]=0, B[i]=0;
    FFT::mul(A,B,C,len);
    for(int i = 0; i < m; ++i){
        cnt[i+1] += C[i+m];
    }

    for(int i = st; i <= ed+1; ++i){
        int le = st==i? l-1 : po[num][i-1];
        int re = ed+1==i? r+1 : po[num][i];
        cdq(le+1, re-1);
    }
}
int main(){
    int T; scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        memset(cnt,0,sizeof(cnt));
        for(int i = 1; i <= n; ++i) po[i].clear();

        for(int i = 1; i <= n; ++i){
            scanf("%d",&a[i]); po[a[i]].push_back(i);
        }
        Build(1,n,1);
        cdq(1,n);

        ll ans = 0;
        for(int i = 1; i <= n; ++i) ans += cnt[i]^i;
        printf("%lld\n",ans);
    }
    return 0;
}