Can you find it
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 584 Accepted Submission(s): 271
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
算法导论上说可以把模运算非正式地与通常的整数运算一样看待，如果仅限于加法，减法，乘法运算。则用这样的非正式模型就足够了。
可以理解为在所有数均mod一个特定数n的情况下，可以进行加减乘运算。
a^(k1*n+b1) + b^(k2*n-k2+1) = 0
a^(k1*n+b1) = -b^(k2*n-k2+1)
a^b1/b^(1-k2) = -(b^k2/a^k1)^n ……………..1
0 = C*(b^k2/a^k1)^n………………………………….2
1+2
得:
a^b1/b^(1-k2) = （C-1）*(b^k2/a^k1)^n
a^b1*b^(k2-1)/(C-1) = (b^k2/a^k1)^n
可知，只有当b^k2 = a^k1，a^b1*b^(k2-1) = (C-1)时（即b^k2/a^k1 = 1）才能满足题意
又当n = 1时，a^(k1+b1) + b = 0
即 a^(k1+b1) + b = C
b = C - a^(k1+b1)
可知b和a是线性的一一对应关系，至此，枚举a，根据公式算出b，判断a，b是否满足条件，满足则输出。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <bitset>
#include <cmath>
using namespace std;
typedef long long LL;
LL fun(LL a,LL b,LL c){
LL ans = 1;
while(b){
if(b&1)ans = ans*a%c;
a= a*a%c;
b>>=1;
}
return ans;
}
int main(){
int CNT = 1;
LL C,k1,b1,k2;
while(scanf("%lld%lld%lld%lld",&C,&k1,&b1,&k2)!=EOF){
printf("Case #%d:\n",CNT++);
int a,b;
bool p = false;
for(a = 1;a<C;a++){
b = C - fun(a,k1+b1,C);
if(C-1 == fun(a,b1,C)*fun(b,k2-1,C)%C && fun(b,k2,C) == fun(a,k1,C)){
p = true;
printf("%d %d\n",a,b);
}
}
if(!p) printf("-1\n");
}
return 0;
}