HDU 5478 Can you find it(快速幂)——2015 ACM/ICPC Asia Regional Shanghai Online

Can you find it

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Given a prime number C(1≤C≤2×105),
and three integers k1, b1, k2 (1≤k1,k2,b1≤109).
   Please find all pairs (a, b) which satisfied the equation ak1⋅n+b1+ bk2⋅n−k2+1 =
0 (mod C)(n = 1, 2, 3, ...).

 

Input

There are multiple test cases (no more than 30). For each test, a single line contains four integers C, k1, b1, k2.

 

Output

First, please output "Case #k: ", k is the number of test case. See sample output for more detail.

Please output all pairs (a, b) in lexicographical order. (1≤a,b<C).
If there is not a pair (a, b), please output -1.

 

Sample Input

23 1 1 2

 

Sample Output

Case #1:
1 22

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

/*********************************************************************/

题意：给你C,k1,k2,b1，按字典序输出满足

的所有(a,b)对

解题思路：因为

对于任意n均满足，故n=1的情况也是符合的，故可得


①

而n=2的情况也是符合的，可得


②

因为①式mod C = 0 ,所以①式乘以一个数mod C 仍为0，不妨①式*

，可得



所以，我们只需遍历一遍a的取值(1~C-1)，利用快速幂计算出

，以及

，再根据式①可以算出b，再用一次快速幂求出

，比较

(mod
C)是否等于

(mod C)即可

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 200001;
const int inf = 1000000000;
const int mod = 2009;
__int64 Quick_Mod(int a, int b, int m)
{
__int64 res = 1,term = a % m;
while(b)
{
if(b & 1) res = (res * term) % m;
term = (term * term) % m;
b >>= 1;
}
return res%m;
}
int main()
{
int C,k1,b1,k2,i,k=1;
__int64 a,b,s;
bool flag;
while(~scanf("%d%d%d%d",&C,&k1,&b1,&k2))
{
flag=false;
printf("Case #%d:\n",k++);
for(i=1;i<C;i++)
{
a=Quick_Mod(i,k1,C);
b=C-Quick_Mod(i,k1+b1,C);
s=Quick_Mod(b,k2,C);
//printf("##%I64d %I64d\n",a,s);
if(a==s)
flag=true,printf("%d %I64d\n",i,b);
}
if(!flag)
puts("-1");
}
return 0;
}

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