题目大意
有n∗m个火炉,排成m层,每层恰好
初始时第一层第i个炉的火力为
energy[i][j]=∑k=1n(energy[i−1][k]+ji)
然后从当前最后一层的有火力的火炉中,把编号最前的s个火炉和最后的
∑i=1s∑j=1s( Ai∗Bj∗(|i−j|+1)2 )
一共进行T操作,每次操作输出最后得到的火力,答案对
Data Constraint
T≤10,n≤1018,m≤5000,s≤30000
题解
先化简一下式子
energy[i][j]=∑k=1n(energy[i−1][k]+ji)=n∗ji+∑k=1nenergy[i−1][k]
令
Sumi=∑j=1nenergy[i][j]
则
energy[i][j]=n∗ji+Sumi−1
又
Sumi=∑j=1nenergy[i][j]=∑j=1n(n∗ji+Sumi−1)=n∗(Sumi−1+∑j=1nji)
然后我们可以O(m2)预处理Si=∑nj=1ji,计算出Summ,再O(slogn)取出A,B的元素。
自然数幂和
计算Si=∑nj=1ji就是求自然数幂和,这里简单讲一种O(n2)的方法。
已知
(a+b)k+1=∑i=0k+1Cik+1∗aibk+1−i
∴(n+1)k+1−nk+1=C1k+1nk+C2k+1nk−1+...+Ckk+1n+1
分别令n=1,2,3,...,n代入后将n个式子相加,得到
移项得
∑i=1nik=1k+1[(n+1)k+1−(C2k+1∑i=1nik−1+...+Ckk+1∑i=1ni+n+1)]
记S(n,k)=∑ni=1ik,原式即为
S(n,k)=1k+1[(n+1)k+1−(C2k+1S(n,k−1)+...+Ckk+1S(n,1)+n+1)]
然后就能递归求解了。
现在取出了A和
∑i=1s∑j=1s( Ai∗Bj∗(|i−j|+1)2 )
(|i−j|+1)2=(i−j)2+1+2|i−j|=i2+j2−2ij+1+2|i−j|
前四项的计算都比较简单,以i2为例:
将式子变形:
∑i=1s∑j=1sAi∗Bj∗i2=∑i=1sAi∗i2∑j=1sBj
这样一来就比较容易计算了,维护对应次数的和即可,其他三项同理。
剩下的绝对值,先讨论i≥j的情况。此时
∑i=1s∑j=1iAi∗Bj∗2(i−j)
拆开括号有两项:2i−2j,以2i为例
∑i=1sAi∗i2∑j=1iBj
维护B的前缀和就很好计算了。
SRC
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std ;
#define N 5000 + 10
#define M 30000 + 10
typedef long long ll ;
const int MO = 1004535809 ;
ll fac[N] , _fac[N] ;
ll S[N] , Sum[N] ;
ll A[M] , B[M] ;
ll sA[M] , sa1[M] , sB[M] , sb1[M] ;
int Case , m , len ;
ll n , ans ;
ll Power( ll x , int k ) {
ll s = 1 ;
x %= MO ;
while ( k ) {
if ( k & 1 ) s = s * x % MO ;
x = x * x % MO ;
k /= 2 ;
}
return s ;
}
void Pre() {
fac[0] = _fac[0] = 1 ;
for (int i = 1 ; i <= 5000 ; i ++ ) {
fac[i] = fac[i-1] * i % MO ;
_fac[i] = Power( fac[i] , MO - 2 ) ;
}
}
ll Calc( int n , int m ) { return fac[n] * _fac[m] % MO * _fac[n-m] % MO ; }
void GetS( int k ) {
if ( k == 1 ) {
S[k] = n % MO * ((n + 1) % MO) % MO * _fac[2] % MO ;
return ;
}
GetS( k - 1 ) ;
S[k] = Power( (n + 1) % MO , k + 1 ) ;
ll sum = (n + 1) % MO ;
for (int i = 2 ; i <= k ; i ++ )
sum = (sum + Calc(k+1,i) * S[k+1-i] % MO) % MO ;
S[k] = (S[k] - sum + MO) % MO ;
S[k] = S[k] * Power( k + 1 , MO - 2 ) % MO ;
}
int main() {
scanf( "%lld%d" , &n , &m ) ;
scanf( "%d" , &Case ) ;
Pre() ;
GetS(m) ;
Sum[1] = S[1] ;
for (int i = 2 ; i <= m ; i ++ )
Sum[i] = (n % MO) * ((Sum[i-1] + S[i]) % MO) % MO ;
int last = 0 ;
while ( Case -- ) {
A[0] = B[0] = ans = 0 ;
scanf( "%d" , &len ) ;
for (ll i = last + 1 ; i <= last + len ; i ++ ) A[++A[0]] = (Sum[m-1] + n % MO * Power(i,m) % MO) % MO ;
for (ll i = n - last - len + 1 ; i <= n - last ; i ++ )
B[++B[0]] = (Sum[m-1] + n % MO * Power(i,m) % MO) % MO ;
ll suma = 0 , suma1 = 0 , suma2 = 0 ;
ll sumb = 0 , sumb1 = 0 , sumb2 = 0 ;
for (int i = 1 ; i <= len ; i ++ ) {
suma = (suma + A[i]) % MO ;
suma1 = (suma1 + A[i] * i % MO) % MO ;
suma2 = (suma2 + A[i] * i % MO * i % MO) % MO ;
sumb = (sumb + B[i]) % MO ;
sumb1 = (sumb1 + B[i] * i % MO) % MO ;
sumb2 = (sumb2 + B[i] * i % MO * i % MO) % MO ;
sA[i] = (sA[i-1] + A[i]) % MO ;
sa1[i] = (sa1[i-1] + 2ll * A[i] % MO * i % MO) % MO ;
sB[i] = (sB[i-1] + B[i]) % MO ;
sb1[i] = (sb1[i-1] + 2ll * B[i] % MO * i % MO) % MO ;
}
ans = (ans + suma * sumb % MO) % MO ;
ans = (ans + suma * sumb2 % MO) % MO ;
ans = (ans + suma2 * sumb % MO) % MO ;
ans = (ans - 2ll * suma1 % MO * sumb1 % MO + MO) % MO ;
for (int i = 1 ; i <= len ; i ++ ) {
ans = (ans + 2ll * i * A[i] % MO * sB[i] % MO) % MO ;
ans = (ans - A[i] * sb1[i] % MO + MO) % MO ;
ll tp = (sB[len] - sB[i] + MO) % MO ;
ans = (ans - 2ll * i * A[i] % MO * tp % MO + MO) % MO ;
tp = (sb1[len] - sb1[i] + MO) % MO ;
ans = (ans + A[i] * tp % MO) % MO ;
}
printf( "%lld\n" , ans ) ;
last += len ;
}
return 0 ;
}
以上.