hdu5478 Can you find it+快速幂
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hdu5478 Can you find it+快速幂 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5478
题面描述:
Can you find it
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1492 Accepted Submission(s): 614
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.
Please output all pairs (a, b) in lexicographical order.
Sample Input
23 1 1 2
Sample Output
Case #1:1 22
Source
2015 ACM/ICPC Asia Regional Shanghai Online
题目大意:
求所有的数对(a,b),使得其满足题目中的公式,但是此题的简单之处就在于,只需要当n=1,2时满足题意即可。
具体解题思路(转):
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
代码实现:
#include <iostream>#include <cstdio>using namespace std;long long quick_mod(long long a,long long b,long long m){ long long ans=1; while(b) { if(b&1) { ans=(ans*a)%m; b--; } b/=2; a=a*a%m; } return ans;}int main(){ long long c; long long k1,b1,k2; int casenum=1; while(scanf("%lld%lld%lld%lld",&c,&k1,&b1,&k2)!=EOF) { int flag=0; long long b=0; long long ans1=0,ans2=0,ans3=0; printf("Case #%d:\n",casenum++); for(long long i=1; i<c; i++) { ans1=quick_mod(i,k1,c)%c; ans2=quick_mod(i,k1+b1,c)%c; b=c-ans2; ans3=quick_mod(b,k2,c)%c; if(ans1==ans3) { flag=1; printf("%lld %lld\n",i,b);//注意:字典序指的是所有数对的字典序 } } if(flag==0) { printf("-1\n"); } } return 0;}
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