zhx's contest(快速幂+快速乘法)
2017-11-27 16:17
246 查看
[align=left]Problem Description[/align]
As one of the most powerful brushes, zhx is required to give his juniors
n
problems.
zhx thinks the ith
problem's difficulty is i.
He wants to arrange these problems in a beautiful way.
zhx defines a sequence {ai}
beautiful if there is an i
that matches two rules below:
1: a1..ai
c522
are monotone decreasing or monotone increasing.
2: ai..an
are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there if the sequence of the problems' difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him the answer module
p.
[align=left]Input[/align]
Multiply test cases(less than
1000).
Seek EOF
as the end of the file.
For each case, there are two integers n
and p
separated by a space in a line. (1≤n,p≤1018)
[align=left]Output[/align]
For each test case, output a single line indicating the answer.
[align=left]Sample Input[/align]
2 233
3 5
[align=left]Sample Output[/align]
2
1
HintIn the first case, both sequence {1, 2} and {2, 1} are legal.
In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1
#include<stdio.h>
typedef long long ll;
ll n,p;
ll qmul(ll a,ll b)
{
ll ans=0;
while(b)
{
if(b%2)
ans=(ans+a)%p;
a=(a+a)%p;
b/=2;
}
return ans;
}
ll qpow(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b%2)
ans=qmul(ans,a)%p;
a=qmul(a,a);
b/=2;
}
return ans;
}
int main()
{
while(scanf("%lld%lld",&n,&p)!=EOF)
{
ll ans;
ans=qpow(2,n)-2;
if(p==1)
{
printf("%d\n",0);
continue;
}
else if(n==1)
{
printf("%d\n",1);
continue;
}
if(ans<0)
ans+=p;
printf("%lld\n",ans);
}
return 0;
}
As one of the most powerful brushes, zhx is required to give his juniors
n
problems.
zhx thinks the ith
problem's difficulty is i.
He wants to arrange these problems in a beautiful way.
zhx defines a sequence {ai}
beautiful if there is an i
that matches two rules below:
1: a1..ai
c522
are monotone decreasing or monotone increasing.
2: ai..an
are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there if the sequence of the problems' difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him the answer module
p.
[align=left]Input[/align]
Multiply test cases(less than
1000).
Seek EOF
as the end of the file.
For each case, there are two integers n
and p
separated by a space in a line. (1≤n,p≤1018)
[align=left]Output[/align]
For each test case, output a single line indicating the answer.
[align=left]Sample Input[/align]
2 233
3 5
[align=left]Sample Output[/align]
2
1
HintIn the first case, both sequence {1, 2} and {2, 1} are legal.
In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1
#include<stdio.h>
typedef long long ll;
ll n,p;
ll qmul(ll a,ll b)
{
ll ans=0;
while(b)
{
if(b%2)
ans=(ans+a)%p;
a=(a+a)%p;
b/=2;
}
return ans;
}
ll qpow(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b%2)
ans=qmul(ans,a)%p;
a=qmul(a,a);
b/=2;
}
return ans;
}
int main()
{
while(scanf("%lld%lld",&n,&p)!=EOF)
{
ll ans;
ans=qpow(2,n)-2;
if(p==1)
{
printf("%d\n",0);
continue;
}
else if(n==1)
{
printf("%d\n",1);
continue;
}
if(ans<0)
ans+=p;
printf("%lld\n",ans);
}
return 0;
}
相关文章推荐
- hdu5187_zhx's contest(快速幂+快速乘法)
- zhx's contest (hdu 5187 快速幂+快速乘法)
- HDU 5187 zhx's contest(快速幂+快速乘法)
- HDU 5187-zhx's contest(快速乘法+快速幂取模)
- HDU 5187 zhx's contest(快速幂乘法)
- hdu 5187 zhx's contest(快速幂,快速乘法,排列组合)
- zhx's contest----快速幂+快速乘
- HDU - 5187 - zhx's contest (快速幂+快速乘)
- BestCoder Round #33(zhx's contest-快速乘法)
- hdu 5187 zhx and contest(快速幂、快速乘法)
- hdu 5187 zhx's contest(快速幂矩阵6)
- zhx's contest BestCoder Round #33(快速乘法+快速幂的原理)
- HDU 5187 zhx's contest(快速乘法)
- zhx's contest (hdu 5187 快速幂+快速乘法)
- 51nod 1137 矩阵乘法(矩阵快速乘法)
- 矩阵乘法快速幂 codevs 1732 Fibonacci数列 2
- python中如何的快速进行除法、乘法运算
- HDOJ-1757A Simple Math Problem(矩阵乘法快速幂)
- poj3070(矩阵快速幂,矩阵乘法)
- codevs1281 矩阵乘法 快速幂 !!!手写乘法取模!!! 练习struct的构造函数和成员函数