1325 - Distributing Chocolates

1325 - Distributing Chocolates

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Time Limit: 4 second(s)

Memory Limit: 32 MB

I have bought n chocolates for my young cousins. Every chocolate is different. So, in the contest I added the problem that how many ways I can distribute the chocolates to my K cousins. I can give more chocolates to some
cousins, and may give no chocolate to some. For example, I have three cousins and I bought 2 chocolates a and b. Then I can distribute them in the following 9 ways:

No.	Cousin 1	Cousin 2	Cousin 3
1	a, b
2		a, b
3			a, b
4	a	b
5	a		b
6		a	b
7	b	a
8	b		a
9		b	a

Now as the result can be large, I asked for the result modulo 100 000 007 (a prime). But after that I found that this problem is easier than I thought. So, I changed the problem a little bit. I will give you the number of my cousins K and
the result modulo 100 000 007. Your task is to find the number of chocolates I have bought. If there are several solutions, you have to find the minimum one.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing two integers K (2 ≤ K ≤ 107) and result (0 ≤ result < 100000007). You can assume that the input data is valid, that means a solution exists.

Output

For each case, print the case number and the minimum possible number of chocolates I have bought.

Sample Input	Output for Sample Input
2 3 9 2 100	Case 1: 2 Case 2: 23502611

 

用到了数论中的baby-step giant-step算法

这个算法就是为了求a的x次方对p取余等于d 求x的值

#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;
const int mod=100000007;
typedef long long LL;
LL Qpow(LL n,LL m){
LL ans=1; LL base=n;
while(m){
if(m&1)
ans=(ans*base)%mod;
base=(base*base)%mod;
m>>=1;
}
return ans;
}
int main(){
int T; scanf("%d",&T); int kcase=1;
while(T--){
int k; int result;
scanf("%d %d",&k,&result);
map<int,int> M;
LL m=ceil(sqrt(mod));
LL x=1;
for(int i=0;i<=m;i++){
M[x]=i;  x=(x*k)%mod;
}
LL t=Qpow(Qpow(k,mod-2),m);
for(int i=0;i<=m;i++){
if(M.find(result)!=M.end()){
printf("Case %d: %d\n",kcase++,m*i+M[result]);
break;
}
result=(result*t)%mod;
}
}
return 0;
}

再贴一个EXGCD的

#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;
const int mod=100000007;
typedef long long LL;
LL p,q;
LL Qpow(LL n,LL m){
LL ans=1; LL base=n;
while(m){
if(m&1)
ans=(ans*base)%mod;
base=(base*base)%mod;
m>>=1;
}
return ans;
}
LL EXGCD(LL a,LL b){
if(b==0){
p=1;q=0;
return a;
}
LL r = EXGCD(b,a%b);
LL t = p;
p = q;
q=t-a/b*q;
return r;

}
int main(){
int T; scanf("%d",&T); int kcase=1;
while(T--){
int k; int result;
scanf("%d %d",&k,&result);
map<LL,int> M;
LL m=ceil(sqrt(mod));
LL x=1;
for(int i=0;i<=m;i++){
M[x]=i;  x=(x*k)%mod;
}
LL mid=1;
LL t=Qpow(k,m);
for(int i=0;i<=m;i++){
LL G=EXGCD(mid,mod);
p=(p*result/G%mod+mod)%(mod/G);
if(M.find(p)!=M.end()){
printf("Case %d: %lld\n",kcase++,m*i+M[p]);
break;
}
mid=(mid*t)%mod;
}
}
return 0;
}

又找到了不互诉的求法的说明
终于稍微有点空了。。 
我们来看这个方程：  


 
a，b，p为常数且在int内。 
注意到这次p可以为合数。 
先来说说p为质数或者合数有什么问题。 
对于a与p互质，那么有a^phi(p)=1(mod p)，对于p是素数phi(p)==p-1，所以x的取值只要在0->n-2之中取就可以了.然而如果p为合数，phi(p) < p-1，这个范围不明确，就不好分块了。而且解是否存在，有几个，也很麻烦。 
因此有extended-baby-step-giant-step算法。 
考虑a与p不互质的情况： 
对于上面的方程，我们可以考虑从x个a中拿出c个a与b和p消去公因子，直到a和p’互质为止。 
一旦互质了，那么方程就是v*a^(x-c)=b’ (mod p’)，v是拿出c个a消去公因子后剩下的东西，b’，p’是消去公因子的b，p。 
这个时候还要求v的逆元，方程变为a^(x-c)=b’*v^(-1) (mod p’)。 
此时就可以用baby-step-giant-step做了，答案为BSGS的答案+c。 
注意：有可能c会大于x，所以必须约去一次就特判一次方程两边（就是v和b’）是不是相等了。如果两边相等那么直接返回c。 
附HDU2815EXBSGS裸题代码（hash表乱写的所以很慢）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
ll a,b,p,ans,hsh[70000][2];
bool ishsh[70000];
void insert(ll x,ll t)
{
ll p=x*12580%70000;
while(ishsh[p])p=(p+1)%70000;
ishsh[p]=1;
hsh[p][0]=x;
hsh[p][1]=t;
}
int query(ll x)
{
ll p=x*12580%70000;
while(hsh[p][0]!=x&&ishsh[p])
p=(p+1)%70000;
if(!ishsh[p])return -1;
return hsh[p][1];
}
ll gcd(ll a, ll b)
{
return b?gcd(b,a%b):a;
}
void exgcd(ll a,ll b,ll &d,ll &x,ll &y)
{
if(!b){x=1;d=a;y=0;}
else
{
exgcd(b,a%b,d,y,x);
y-=a/b*x;
}
}
ll inv(ll a,ll p)
{
ll x,y,d;
exgcd(a,p,d,x,y);
return (x+p)%p;
}
ll BSGS(ll a,ll b,ll p)
{
ll m=ceil(sqrt(p)),d=1,val=1,gcd,x,y,t;
for(int i=0;i<m;++i)
{
insert(val,i);
val=val*a%p;
}
for(int i=0;i<m;++i)
{
exgcd(d,p,gcd,x,y);
x=(b/gcd*x%p+p)%(p/gcd);
t=query(x);
if(t!=-1)return i*m+t;
d=d*val%p;
}
return -1;
}
ll EXBSGS(ll a,ll b,ll p)
{
ll t,c=0,v=1;
while((t=gcd(a,p))!=1)
{
if(b%t)return -1;
p/=t;//p一定要先除
b/=t;
v=v*a/t%p;//因为这里%p是约去后的p
++c;
if(b==v)return c;//特判
}
b=b*inv(v,p)%p;
ll ret=BSGS(a,b,p);
return ret!=-1?ret+c:ret;
}
int main()
{
while(~scanf("%I64d%I64d%I64d",&a,&p,&b))
{
memset(hsh,0,sizeof hsh);
memset(ishsh,0,sizeof ishsh);
if(b>=p)puts("Orz,I can’t find D!");//引号是全角的- -
else if(b==0)puts("0");
else if((ans=EXBSGS(a,b,p))==-1)puts("Orz,I can’t find D!");
else printf("%I64d\n",ans);
}
}