例题10-1 巨大的斐波那契数！ UVa11582

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1.题目描述：点击打开链接

2.解题思路：由于是模运算，因此整个序列肯定会出现重复序列，事先可以枚举周期T，然后利用pow_mod的算法快速计算a^b(mod T)的结果p，返回F[p]即可。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef unsigned long long LL;const int N = 1010;LL a, b;int n;int memo[N];int T(int n) //计算周期T{int& res = memo[n];if (res)return res;int prev = 0, curr = 1, succ;res = 0;for (;;) {res++;succ = (prev + curr) % n;prev = curr;curr = succ;if (prev == 0 && curr == 1)return res;}}int pow_mod(LL a, LL k, int mod) //计算a^k模mod的结果{a %= mod;LL res = 1;while (k) {if (k & 1) res = (res*a) % mod;a = (a*a) % mod;k >>= 1;}return (int)res;}int F(int n, int mod) //计算第n项{int prev = 0, curr = 1, succ;while (n--) {succ = (prev + curr) % mod;prev = curr;curr = succ;}return prev;}int main() {//freopen("test.txt", "r", stdin);int t;cin >> t;while (t--) {cin >> a >> b >> n;if (n == 1) {cout << 0 << endl;continue;}cout << F(pow_mod(a, b, T(n)), n) << endl;}return 0;}

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