HDU 5478 Can you find it(数学归纳法 + 快速幂)——2015 ACM/ICPC Asia Regional Shanghai Online
2016-10-25 21:06
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Total Submission(s): 1487 Accepted Submission(s): 614
[align=left]Problem Description[/align] 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, …).
[align=left]Input[/align] There are multiple test cases (no more than 30). For each test, a single line contains four integers C, k1, b1, k2.
[align=left]Output[/align] 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.
[align=left]Sample Input[/align]
[align=left]Sample Output[/align]
题目大意:
给你四个数 C,k1,b1,k2,让你求上面的公式满足所有的 n成立的 (a,b) 的对,并且将 (a,b) 按照 a 的字典序输出,如果没有输出 −1。
解题思路:
因为对所有的 n 都满足所以我们取两个特殊的情况也就是 n==1 和 n==2 的情况,然后解出一对 (a,b) ,如果这个 (a,b) 满足的话,对于所有的 n 也是满足的,可以根据数学归纳法证明一下,所以现在就是暴力找 a 就行了。
My Code:
Can you find it
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1487 Accepted Submission(s): 614
[align=left]Problem Description[/align] 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, …).
[align=left]Input[/align] There are multiple test cases (no more than 30). For each test, a single line contains four integers C, k1, b1, k2.
[align=left]Output[/align] 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.
[align=left]Sample Input[/align]
23 1 1 2
[align=left]Sample Output[/align]
Case #1: 1 22
题目大意:
给你四个数 C,k1,b1,k2,让你求上面的公式满足所有的 n成立的 (a,b) 的对,并且将 (a,b) 按照 a 的字典序输出,如果没有输出 −1。
解题思路:
因为对所有的 n 都满足所以我们取两个特殊的情况也就是 n==1 和 n==2 的情况,然后解出一对 (a,b) ,如果这个 (a,b) 满足的话,对于所有的 n 也是满足的,可以根据数学归纳法证明一下,所以现在就是暴力找 a 就行了。
My Code:
#include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; typedef long long LL; LL Pow(LL a, LL b, LL MOD){ LL ans = 1; while(b){ if(b&1) ans = (ans*a)%MOD; b>>=1; a = (a*a)%MOD; } return ans; } int main() { LL C, k1, b1, k2; int cas = 1; while(cin>>C>>k1>>b1>>k2){ LL tp = k1+b1; int ok = 0; printf("Case #%d:\n",cas++);; for(LL a=1; a<C; a++){ LL tmp = Pow(a, tp , C); LL b = C - tmp; if((Pow(a,tp+k1, C)+Pow(b, k2+1, C))%C == 0){ ok = 1; printf("%I64d %I64d\n",a,b); } } if(!ok) puts("-1"); } return 0; }
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