UVALive 7361 Immortal Porpoises （矩阵快速幂）

来源：互联网发布：七天网络查询分数编辑：程序博客网时间：2024/06/10 12:56

大体题意：

让你求n个斐波那契数列，对1e9取模！

思路：

很明显矩阵快速幂求矩阵1 1

1 0的n 次方即可！

但是好久没写了，写了很久很久!!= = !

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<cstdlib>#include<vector>#include<set>#include<map>#include<stack>#include<cctype>#include<string>#include<iostream>#include<queue>#define fout freopen("out.txt","w","stdout");#define mr make_pair#define fi first#define se secondusing namespace std;typedef long long ll;typedef unsigned long long LLU;const int maxn = 1000 + 5;const int mod = 1000000000;const double eps = 1e-8;const int inf = 0x3f3f3f3f;struct Mar{    ll a[2][2];    void clear(){        memset(a,0,sizeof a);    }     bool operator = (Mar  rhs)  {        a[0][0] = rhs.a[0][0];        a[0][1] = rhs.a[0][1];        a[1][0] = rhs.a[1][0];        a[1][1] = rhs.a[1][1];        return 1;    }}unit;Mar mul(Mar m1,Mar m2){    Mar tmp;    tmp.clear();    tmp.a[0][0] = (m1.a[0][0]%mod*m2.a[0][0]%mod + m1.a[0][1]%mod * m2.a[1][0]%mod)%mod;    tmp.a[0][1] = (m1.a[0][0]%mod*m2.a[0][1]%mod + m1.a[0][1]%mod * m2.a[1][1]%mod)%mod;    tmp.a[1][0] = (m1.a[1][0]%mod*m2.a[0][0]%mod + m1.a[1][1]%mod * m2.a[1][0]%mod)%mod;    tmp.a[1][1] = (m1.a[1][0]%mod*m2.a[0][1]%mod + m1.a[1][1]%mod * m2.a[1][1]%mod)%mod;    return tmp;}Mar my_pow(Mar a,ll n,ll id){    Mar ans = unit;    while(n){        if (n & 1)            ans = mul(ans,a);        n/=2;        a = mul(a,a);    }    return ans;}ll gcd(ll a,ll b){    return !b ? a : gcd(b,a%b);}ll lcm(ll a,ll b){    return a*b/gcd(a,b);}int main(){    unit.a[0][0] = 1;    unit.a[0][1] = 0;    unit.a[1][0] = 0;    unit.a[1][1] = 1;    int T;    scanf("%d",&T);    while(T--){        ll k,n;        Mar tmp;        tmp.a[0][0] = 1;        tmp.a[0][1] = 1;        tmp.a[1][0] = 1;        tmp.a[1][1] = 0;        Mar tt = mul(tmp,tmp);        scanf("%lld %lld",&k,&n);        Mar ans = my_pow(tmp,n,k);        printf("%lld %lld\n",k,ans.a[1][0]);    }    return 0;}

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