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yandex资格赛C(dp->快速幂)


这场的bc都不太容易切啊 。。b的话还是考逻辑想象。。思维度没有。。

然后这个c还是蛮有意思。。

考虑一维情况很容易想到dp方程

设扫到第i个数前缀和余数为j的方案为d[i][j]

d[i][j]=sum(d[i-1][k])  k=0-n-1

       =n*d[i-1][j]

因为对上个余数k肯定能在1-n找到数使余数为 j。。所以余数为几基本都一样。。可以简化为:

d[i]*=n*d[i-1]*

求d[n][0]直接n^(n-1)即可。。

推广到二维。。

设扫到第i列j行前缀和余数为tj为d[i][t1][t2][t3]...[tn]

d[i][t1][t2][t3]...[tn]=sum(d[i-1][t1'][t2']..[tn'])

d[i-1]里面有n^n个状态供转移,但确不是d[i]*=n^n*d[i-1]*,因为多了个列和被n整除的情况,取数有限制,并不是所有的状态都能转移。。

所以该怎么办?换个想法。。如果随意取前n-1个数,那么第n个数一定是存在且确定的。。那这样我们将方程的维数降1即d[i][t1][t2]...[tn-1]。。这样n^(n-1)个状态就能完全转移。。。

d[i]*=n^(n-1)*d[i-1]*

就变成求n^(n-1)^(n-1)=n^((n-1)^2)了。。

然后直接快速幂。。。







/**
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*        ┃       ┃  
*        ┃   ━    ┃
*        ┃ >   < ┃
*        ┃       ┃
*        ┃... ⌒ ...  ┃
*        ┃       ┃
*        ┗━┓   ┏━┛
*          ┃   ┃ Code is far away from bug with the animal protecting          
*          ┃   ┃ 神兽保佑,代码无bug
*          ┃   ┃           
*          ┃   ┃       
*          ┃   ┃
*          ┃   ┃           
*          ┃   ┗━━━┓
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*          ┗┓┓┏━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 205
#define nm 100005
#define pi 3.1415926535897931
using namespace std;
const ll inf=1000000007;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}









ll n;
ll pow(ll x,ll t){return t==0?1:pow(sqr(x)%inf,t>>1)*(t&1?x:1)%inf;}

int main(){
n=read();
printf("%lld\n",pow(n,sqr(n-1)));
}




防止找不到原题贴个链接:

​​https://contest.yandex.com/algorithm2018/contest/7491/problems/C/​​



Beautiful Tables

Time limit

1 second

Memory limit

512Mb

Input

standard input or input.txt

Output

standard output or output.txt


Victor usually spends free time with reading books, solving riddles and puzzles.

Yesterday he decided to come up with his own puzzle. One should fill the table of size n × n with integers from 1 to n in a such way that the sum of the numbers in each row and the sum of numbers in each column is divisible by n. Each integer from 1 to n

Help Victor to determine the number of distinct tables satisfying the requirements of the puzzle. Two table are considered to be distinct if they differ in at least one cell. As the number Victor wants to compute may be pretty big, you only need to find its remainder modulo 109 + 7.


Input format


The only input line contains a single integer n (1 ≤ n ≤ 1000), the number of row and the number of columns of the table.


Output format


Print the number of appropriate tables modulo 1 000 000 007.


Sample 1

Input

Output

2

2

Sample 2

Input

Output

3

81


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