5069. 蛋糕

题目大意

求在 n×n 的网格图中彼此可见的点之间的线段总数。

Data Constraint
n≤109

题解

题目要求的就是

2[2×n2∑i=1nφ(i)−3n∑i=1nφ(i)×i+∑i=1nφ(i)×i2]

然后就可以考虑用杜教筛来计算。

1.计算 S(n)=∑ni=1φ(i)
对于 φ ，具有 ∑d|iφ(d)=i 的性质。
那么

∑i=1n∑d|iφ(d)=n(n+1)2

先枚举约数，转化一下

∑i=1n∑j=1⌊ni⌋φ(j)=n(n+1)2

然后套路一波

S(n)+∑i=2nS(⌊ni⌋)=n(n+1)2

2.计算 S(n)=∑ni=1φ(i)×i
因为 ∑d|iφ(d)=i ，所以 ∑d|iφ(d)×d×id=i2

∑i=1n∑d|iφ(d)×d×id=∑i=1ni2=n(n+1)(2n+1)6

∑i=1n∑j=1⌊ni⌋φ(j)×i×j=n(n+1)(2n+1)6

S(n)+∑i=2ni×S(⌊ni⌋)=n(n+1)(2n+1)6

3.计算 S(n)=∑ni=1φ(i)×i2 与第二种类似。
最后化出来

S(n)+∑i=2ni2×S(⌊ni⌋)=(n(n+1)2)2

SRC

#include
#include
#include
#include
#include
#include
using namespace std ;

#define N 10000000 + 10
#define M 40000 + 10
typedef long long ll ;
const int MAXN = 5e6 ;

ll Q[3][M] ;

bool flag[N] ;
int Pri[N] , phi[N] ;
ll Sum[3][N] ;
int n , MO , Block ;

void Pre() {
    phi[1] = 1 ;
    for (int i = 2 ; i <= MAXN ; i ++ ) {
        if ( !flag[i] ) {
            Pri[++Pri[0]] = i ;
            phi[i] = i - 1 ;
        }
        for (int j = 1 ; j <= Pri[0] ; j ++ ) {
            if ( 1ll * i * Pri[j] > MAXN ) break ;
            flag[i*Pri[j]] = 1 ;
            if ( i % Pri[j] == 0 ) {
                phi[i*Pri[j]] = phi[i] * Pri[j] ;
                break ;
            }
            phi[i*Pri[j]] = phi[i] * (Pri[j] - 1) ;
        }
    }
    for (int i = 1 ; i <= MAXN ; i ++ ) {
        Sum[0][i] = (Sum[0][i-1] + phi[i]) % MO ;
        Sum[1][i] = (Sum[1][i-1] + (ll)i * phi[i] % MO) % MO ;
        Sum[2][i] = (Sum[2][i-1] + (ll)i * i % MO * phi[i] % MO) % MO ;
    }
}

ll Calc( int k , ll n ) {
    if ( k == -1 ) return n % MO ;
    ll ret ;
    if ( k != 1 ) ret = (n * (n + 1) / 2) % MO ;
    else {
        ll ret1 = n , ret2 = n + 1 , ret3 = 2 * n + 1 ;
        if ( ret1 % 2 == 0 ) ret1 /= 2 ;
        else if ( ret2 % 2 == 0 ) ret2 /= 2 ;
        else if ( ret3 % 2 == 0 ) ret3 /= 2 ;
        if ( ret1 % 3 == 0 ) ret1 /= 3 ;
        else if ( ret2 % 3 == 0 ) ret2 /= 3 ;
        else if ( ret3 % 3 == 0 ) ret3 /= 3 ;
        ret1 = ret1 * ret2 >= MO ? ret1 * ret2 % MO : ret1 * ret2 ;
        ret1 = ret1 * ret3 >= MO ? ret1 * ret3 % MO : ret1 * ret3 ;
        return ret1 ;
    }
    if ( k == 2 ) ret = ret * ret >= MO ? ret * ret % MO : ret * ret ;
    return ret ;
}

ll S( int x , int k ) {
    if ( x <= MAXN ) return Sum[k][x] ;
    int t = x <= Block ? x : n / x + Block ;
    if ( Q[k][t] ) return Q[k][t] ;
    ll ret = Calc( k , x ) ;
    for (int i = 2 ; i <= x ; i ++ ) {
        int r = x / (x / i) ;
        ret = (ret - (Calc( k - 1 , r ) - Calc( k - 1 , i - 1 )) % MO * S(x/i,k) % MO) % MO ;
        i = r ;
    }
    Q[k][t] = ret ;
    return ret ;
}

int main() {
    freopen( "cake.in" , "r" , stdin ) ;
    freopen( "cake.out" , "w" , stdout ) ;
    scanf( "%d%d" , &n , &MO ) ;
    Block = sqrt(n) ;
    Pre() ;
    ll ans = (ll)n * n % MO * 2ll % MO * S(n,0) % MO ;
    ans = (ans - 3ll * n % MO * S(n,1) % MO) % MO ;
    ans = (ans + S(n,2)) % MO ;
    ans = 2ll * ans % MO ;
    printf( "%lld\n" , (ans + MO) % MO ) ;
    return 0 ;
}

以上.