UVA 11582 巨大数的斐波那契数列 (大数取模,幂取模,模的计算方
发布时间:2021-02-20 21:03:39 所属栏目:大数据 来源:网络整理
导读:Problem F: Colossal Fibonacci Numbers! The? i 'th Fibonacci number? f?(i) ?is recursively defined 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 seq
|
Problem F: Colossal Fibonacci Numbers!
The?i'th Fibonacci number?f?(i)?is recursively defined in the following way:
Your task is to compute some values of this sequence. Input begins with an integer?t?≤?10,000,the number of test cases. Each test case consists of three integers?a,b,nwhere 0?≤?a,b?<?264?(a?and?b?will not both be zero) and 1?≤?n?≤?1000. For each test case,output a single line containing the remainder of?f?(ab)?upon division by?n. Sample input3 1 1 2 2 3 1000 18446744073709551615 18446744073709551615 1000 Sample output1 21 250 一.大数取模,幂取模,模的计算方法1. (a+b)%n=(a%n+b%n)%n 2.(a-b)%n=(a%n-b%n+n)%n 3.(a*b)%n=(a%n*b%n)%n一.问题分析及解决 想啊,这个问题明明以前都做过很类似的问题的,F[i]%n,那么应该很自然的去想F[i]%n=(F[i-1]%n+F[i-2]%n)%n,接下来应该就会发现要求的东西只要知道前两个数的模就好了,那就很自然的转化到以前的找规律问题上来 #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef unsigned long long ull;
const int maxn=1000+10;
ull F[maxn*maxn];
int power(ull a,ull b,int t) {
ull result = 1;
while(b) {
if(b%2) {
result = (a*result)%t;
}
a = (a*a)%t;
b/=2;
}
return result;
}
int main()
{
freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);
int T;
cin>>T;
while(T--)
{
ull a,n,l;
int flag=1;
cin>>a>>b>>n;
F[0]=0;
F[1]=1;
for(ull i=2;i<=n*n;i++)
{
F[i]=(F[i-1]%n+F[i-2]%n)%n;
if(F[i]==F[1]&&F[i-1]==F[0])
{
l=i-1;
break;
}
}
if(a == 0 || n == 1) printf("0n");
else cout<<F[power(a%l,l)]<<endl;
}
return 0;
}
(编辑:邯郸站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |



