版权声明：本人初三蒟蒻一只，欢迎各位大佬来点评。 https://blog.csdn.net/HiChocolate/article/details/82952339
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-1\equiv 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+1\equiv 1(mod p)
2 p 2 − 4 p + 1 ≡ 1 ( m o d p )
1
≡
1
(
m
o
d
p
)
1\equiv1(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.
