一行两个整数k，p。
一行一个整数表示答案。
1 3
6
对于20%的数据：k ∗ p < = 1 0 5 k*p<=10^5 k ∗ p < = 1 0 5 k = 1 k=1 k = 1 k ∗ p < = 1 0 9 k*p<=10^9 k ∗ p < = 1 0 9 k , p < = 1 0 9 k,p<=10^9 k , p < = 1 0 9
这题一看就是一道不友好的题。然而并不会做 ∑ i = 1 k ∗ p i 2 ∗ p − 1 m o d p 2 sum_{i=1}^{k*p}i^{2*p-1} mod p^2 ∑ i = 1 k ∗ p i 2 ∗ p − 1 m o d p 2 ( a + b ) n = ∑ i = 0 n C n i ∗ a n − i ∗ b i (a+b)^n=sum_{i=0}^{n}C_{n}^{i}*a^{n-i}*b^i ( a + b ) n = ∑ i = 0 n C n i ∗ a n − i ∗ b i
然后我们要证明：∑ i = 1 p − 1 i 2 p − 1 ≡ p ( p + 1 ) / 2 ( m o d p 2 ) sum_{i=1}^{p-1}i^{2p-1}equiv p(p+1)/2(mod p^2) ∑ i = 1 p − 1 i 2 p − 1 ≡ p ( p + 1 ) / 2 ( m o d p 2 ) ∑ i = 1 p − 1 i 2 p − 1 + ∑ i = 1 p − 1 i 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}i^{2p-1}+sum_{i=1}^{p-1}i^{2p-1}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 i 2 p − 1 + ∑ i = 1 p − 1 i 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 )
∑ i = 1 p − 1 i 2 p − 1 + ∑ i = 1 p − 1 ( p − i ) 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}i^{2p-1}+sum_{i=1}^{p-1}{(p-i)}^{2p-1}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 i 2 p − 1 + ∑ i = 1 p − 1 ( p − i ) 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 )
∑ i = 1 p − 1 i 2 p − 1 + ( p − i ) 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}{i}^{2p-1}+{(p-i)}^{2p-1}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 i 2 p − 1 + ( p − i ) 2 p − 1 ≡ p ( p + 1 ) ( m o d p 2 )
∑ i = 1 p − 1 i 2 ∗ p − 1 − ∑ j = 0 2 p − 1 C 2 p − 1 j ∗ ( − i ) 2 p − 1 − j ∗ p j ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}{i}^{2*p-1}-sum_{j=0}^{2p-1}C_{2p-1}^j*(-i)^{2p-1-j}*p^{j}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 i 2 ∗ p − 1 − ∑ j = 0 2 p − 1 C 2 p − 1 j ∗ ( − i ) 2 p − 1 − j ∗ p j ≡ p ( p + 1 ) ( m o d p 2 )
由于当j>=2时，后面的式子是没有用的“∑ j = 0 2 p − 1 C 2 p − 1 j ∗ ( − i ) 2 p − 1 − j ∗ p j sum_{j=0}^{2p-1}C_{2p-1}^j*(-i)^{2p-1-j}*p^{j} ∑ j = 0 2 p − 1 C 2 p − 1 j ∗ ( − i ) 2 p − 1 − j ∗ p j S = − 1 ∗ 1 ∗ i 2 p − 1 S=-1*1*i^{2p-1} S = − 1 ∗ 1 ∗ i 2 p − 1 S = ( 2 p − 1 ) ∗ i 2 p − 2 ∗ p S=(2p-1)*i^{2p-2}*p S = ( 2 p − 1 ) ∗ i 2 p − 2 ∗ p ∑ i = 1 p − 1 i 2 p − 1 − i 2 p − 1 + ( 2 p − 1 ) ∗ p ∗ i 2 p − 2 ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}{i}^{2p-1}-i^{2p-1}+(2p-1)*p*i^{2p-2}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 i 2 p − 1 − i 2 p − 1 + ( 2 p − 1 ) ∗ p ∗ i 2 p − 2 ≡ p ( p + 1 ) ( m o d p 2 )
∑ i = 1 p − 1 ( 2 p − 1 ) ∗ p ∗ i 2 p − 2 ≡ p ( p + 1 ) ( m o d p 2 ) sum_{i=1}^{p-1}(2p-1)*p*i^{2p-2}equiv p(p+1)(modp^2) ∑ i = 1 p − 1 ( 2 p − 1 ) ∗ p ∗ i 2 p − 2 ≡ p ( p + 1 ) ( m o d p 2 )
∑ i = 1 p − 1 ( 2 p − 1 ) ∗ i 2 p − 2 ≡ p + 1 ( m o d p ) sum_{i=1}^{p-1}(2p-1)*i^{2p-2}equiv p+1(modp) ∑ i = 1 p − 1 ( 2 p − 1 ) ∗ i 2 p − 2 ≡ p + 1 ( m o d p )
因为费马小定理：a p − 1 ≡ 1 ( m o d p ) ( p 为 质 数 ) a^{p-1}equiv 1 (mod p)(p为质数) a p − 1 ≡ 1 ( m o d p ) ( p 为 质 数 )
所以i 2 p − 2 = i 2 ( p − 1 ) = ( i 2 ) p − 1 ≡ 1 ( m o d p ) i^{2p-2}=i^{2(p-1)}={(i^2)}^{p-1}equiv 1(modp) i 2 p − 2 = i 2 ( p − 1 ) = ( i 2 ) p − 1 ≡ 1 ( m o d p )
所以原来就变成了：∑ i = 1 p − 1 2 p − 1 ≡ p + 1 ( m o d p ) sum_{i=1}^{p-1}2p-1equiv p+1(modp) ∑ i = 1 p − 1 2 p − 1 ≡ p + 1 ( m o d p )
( p − 1 ) ( 2 p − 1 ) ≡ p + 1 ( m o d p ) (p-1)(2p-1)equiv p+1(mod p) ( p − 1 ) ( 2 p − 1 ) ≡ p + 1 ( m o d p )
2 p 2 − 4 p + 1 ≡ 1 ( m o d p ) 2p^2-4p+1equiv 1(mod p) 2 p 2 − 4 p + 1 ≡ 1 ( m o d p ) 1 ≡ 1 ( m o d p ) 1equiv1(mod p) 1 ≡ 1 ( m o d p ) ∑ i = 1 k p i 2 p − 1 m o d p 2 sum_{i=1}^{kp}i^{2p-1} mod p^2 ∑ i = 1 k p i 2 p − 1 m o d p 2
∑ i = 1 k ∑ j = 1 p [ ( i − 1 ) ∗ p + j ] 2 p − 1 m o d p 2 sum_{i=1}^{k}sum_{j=1}^{p}[(i-1)*p+j]^{2p-1} mod p^2 ∑ i = 1 k ∑ j = 1 p [ ( i − 1 ) ∗ p + j ] 2 p − 1 m o d p 2 m o d p 2 modp^2 m o d p 2 ∑ i = 0 k − 1 ∑ j = 1 p − 1 ( i ∗ p + j ) 2 p − 1 m o d p 2 sum_{i=0}^{k-1}sum_{j=1}^{p-1}(i*p+j)^{2p-1} mod p^2 ∑ i = 0 k − 1 ∑ j = 1 p − 1 ( i ∗ p + j ) 2 p − 1 m o d p 2 ∑ i = 0 k − 1 ∑ j = 1 p − 1 [ j 2 p − 1 + ( 2 p − 1 ) ∗ i ∗ p ∗ j 2 p − 2 ] m o d p 2 sum_{i=0}^{k-1}sum_{j=1}^{p-1}[j^{2p-1}+(2p-1)*i*p*j^{2p-2}] mod p^2 ∑ i = 0 k − 1 ∑ j = 1 p − 1 [ j 2 p − 1 + ( 2 p − 1 ) ∗ i ∗ p ∗ j 2 p − 2 ] m o d p 2 ∑ i = 0 k − 1 ∑ j = 1 p − 1 j 2 p − 1 + ∑ i = 0 k − 1 ∗ i ∗ ∑ j = 1 p − 1 ( 2 p − 1 ) ∗ j 2 p − 2 ∗ p ( m o d p 2 ) sum_{i=0}^{k-1}sum_{j=1}^{p-1}j^{2p-1}+sum_{i=0}^{k-1}*i*sum_{j=1}^{p-1}(2p-1)*j^{2p-2}*p (mod p^2) ∑ i = 0 k − 1 ∑ j = 1 p − 1 j 2 p − 1 + ∑ i = 0 k − 1 ∗ i ∗ ∑ j = 1 p − 1 ( 2 p − 1 ) ∗ j 2 p − 2 ∗ p ( m o d p 2 ) ∑ i = 0 k − 1 p ( p + 1 ) / 2 = k p ( p + 1 ) / 2 sum_{i=0}^{k-1}p(p+1)/2=kp(p+1)/2 ∑ i = 0 k − 1 p ( p + 1 ) / 2 = k p ( p + 1 ) / 2 ∑ i = 0 k − 1 ∗ i ∗ p ∗ ( p + 1 ) = k ∗ ( k − 1 ) ∗ p ∗ ( p + 1 ) / 2 sum_{i=0}^{k-1}*i*p*(p+1)=k*(k-1)*p*(p+1)/2 ∑ i = 0 k − 1 ∗ i ∗ p ∗ ( p + 1 ) = k ∗ ( k − 1 ) ∗ p ∗ ( p + 1 ) / 2 k p ( p + 1 ) / 2 + k ∗ ( k − 1 ) ∗ p ∗ ( p + 1 ) / 2 ( m o d p 2 ) kp(p+1)/2+k*(k-1)*p*(p+1)/2( mod p^2) k p ( p + 1 ) / 2 + k ∗ ( k − 1 ) ∗ p ∗ ( p + 1 ) / 2 ( m o d p 2 ) k 2 ∗ p ∗ ( p + 1 ) / 2 ( m o d p 2 ) k^2*p*(p+1)/2(mod p^2) k 2 ∗ p ∗ ( p + 1 ) / 2 ( m o d p 2 )
var
i,j,k,l,n,m:longint;
p,q,ans,answer,mo:int64;
function tsm(a,b:int64):int64;
var
i,j,k,l:int64;
begin
l:=0;
while b>0 do
begin
if b and 1>0 then l:=(l+a) mod mo;
b:=b div 2;
a:=(a*2) mod mo;
end;
exit(l);
end;
begin
readln(q,p);
mo:=p*p;
writeln(tsm(tsm(tsm(q,q),p),(p+1) div 2));
end.
