HDU 5478 Can you find it 随机化 数学

题意

问你有多少对数，满足a^(k1⋅n+b1) + b^(k2⋅n−k2+1) = 0 (mod C)

题解：

首先你要知道，对于每个a只有唯一对应的b可以满足这个式子，因为当n=1的时候，a^(k1+b1)+b = kk*C

由于b是小于c的，所以只有一个

所以我们可以求出b来，然后我们怎么check这个b究竟是不是呢？

随机化10个数，然后随便check就好了

//qscqesze#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <cmath>#include <cstring>#include <ctime>#include <iostream>#include <algorithm>#include <set>#include <bitset>#include <vector>#include <sstream>#include <queue>#include <typeinfo>#include <fstream>#include <map>#include <stack>typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 100006#define mod 1000000007#define eps 1e-9#define e exp(1.0)#define PI acos(-1)const double EP  = 1E-10 ;int Num;//const int inf=0x7fffffff;const ll inf=999999999;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//*************************************************************************************ll fMul(int m,ll n,int k){    ll cc = m;    ll b = 1;    while (n > 0)    {          if (n & 1LL)            {                b = (b*cc);                if(b>=k)                    b%=k;            }          n = n >> 1LL ;          cc = (cc*cc)%k;          if(cc>=k)cc%=k;    }    return b;}int main(){    //freopen("out.txt","r",stdin);    //freopen("out2.txt","w",stdout);    srand(time(NULL));    int tot = 1;    int c ,k1 ,b1 ,k2;    while(scanf("%d%d%d%d",&c,&k1,&b1,&k2)!=EOF)    {        printf("Case #%d:\n",tot++);        int flag1 = 0;        for(int i=1;i<c;i++)        {            int j=c-fMul(i,k1*1+b1,c);            int flag = 1;            for(int k=1;k<=15;k++)            {                ll tt = rand()%c+1;                ll ttt1 = k1, ttt2 = k2,ttt3 = b1;                if((fMul(i,ttt1*tt+ttt3,c)+fMul(j,ttt2*tt-ttt2+1LL,c))%c!=0)                {                    flag = 0;                    break;                }            }            if(flag)            {                printf("%d %d\n",i,j);                flag1=1;            }        }        if(!flag1)            printf("-1\n");    }}