巨大的斐波那契数列
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The i’th Fibonacci number f(i) is recursively de?ned in the following way: • f(0) = 0 and f(1) = 1 • f(i + 2) = f(i + 1) + f(i) for every i ≥ 0 Your task is to compute some values of this sequence.
Input Input begins with an integer t ≤ 10,000, the number of test cases. Each test case consists of three integers a, b, n where 0 ≤ a,b < 264 (a and b will not both be zero) and 1 ≤ n ≤ 1000.
Output
For each test case, output a single line containing the remainder of f(ab) upon division by n.
Sample Input
3 1 1 2 2 3 1000 18446744073709551615 18446744073709551615 1000
Sample Output
1 21 250
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> typedef unsigned long long ll; using namespace std; const int maxn=1000+10; ll a,b; int f[maxn*maxn],n,M; int pow(ll a,ll p,int Mod) { int ret=1; while(p) { if(p & 1)ret*=a,ret%=Mod; a*=a;a%=Mod; p>>=1; } return ret; } inline void solve() { cin>>a>>b>>n; if(n==1||!a){printf("0\n");return ;} f[1]=1,f[2]=1; for(int i=3;i<=n*n+10;i++) { f[i]=f[i-1]+f[i-2];f[i]%=n; if(f[i]==f[2]&&f[i-1]==f[1]) {M=i-2;break;} } int k=pow(a%M,b,M); printf("%d\n",f[k]); } int main() { int T;cin>>T; while(T--) solve(); return 0; }
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