[bsoj1855] Fibonacci数列

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题目描述

　　Fibonacci数列是大家再熟悉不过的东东了.
　　由于吃完饭后没有什么事做,ChengYu决定计算Fibonacci数列的第n项来帮助消化,现在他希望你也做同样的事. ^_^ Fibonacci数列增长非常快,为了节约宝贵的纸张,所以只要计算答案除以1729的余数即可.

输入格式

多组数据,每行一个正整数n(n<=3456789012).

输出格式

对于每个输入,输出结果.

样例数据

样例输入

1
2
3

样例输出

1
1
2

题目分析

弱智题目，矩阵加速递推，也可以找循环节，略过。

源代码

找循环节

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>using namespace std;long long f[5005],n;int Fibonacci(int n) {    f[1]=0;f[2]=1;    for(int i=2;i<=n;i++) {        f[0]=f[1];        f[1]=f[2];        f[2]=(f[1]+f[0])%1729;    }    return f[2]%1729;}int main() {    while(cin>>n) {        n%=1008;        if(n==0)cout<<Fibonacci(1008)<<endl;        else cout<<Fibonacci(n)<<endl;    }    return 0;}

矩阵快速幂

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>using namespace std;inline const int Get_Int() {    int num=0,bj=1;    char x=getchar();    while(x<'0'||x>'9') {        if(x=='-')bj=-1;        x=getchar();    }    while(x>='0'&&x<='9') {        num=num*10+x-'0';        x=getchar();    }    return num*bj;}typedef long long LL;const int maxn=5,mod=1729;struct Matrix {    LL n,m,a[maxn][maxn];    Matrix(LL n,LL m) {        this->n=n;        this->m=m;        memset(a,0,sizeof(a));    }    Matrix(LL n,LL m,char E) { //单位矩阵        this->n=n;        this->m=m;        memset(a,0,sizeof(a));        for(int i=1; i<=n; i++)a[i][i]=1;    }    void init(LL n,LL m) {        this->n=n;        this->m=m;        memset(a,0,sizeof(a));    }    LL* operator [] (const LL x) {        return a[x];    }    Matrix operator = (Matrix b) {        init(b.n,b.m);        for(int i=1; i<=b.n; i++)            for(int j=1; j<=b.m; j++)                a[i][j]=b[i][j];    }    Matrix operator + (Matrix& b) {        for(int i=1; i<=b.n; i++)            for(int j=1; j<=b.m; j++)                a[i][j]+=b[i][j];    }    void operator += (Matrix& b) {        *this=*this+b;    }    Matrix operator - (Matrix& b) {        for(int i=1; i<=b.n; i++)            for(int j=1; j<=b.m; j++)                a[i][j]-=b[i][j];    }    void operator -= (Matrix& b) {        *this=*this-b;    }    Matrix operator * (LL k) {        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                a[i][j]*=k;    }    void operator *= (LL k) {        *this=*this*k;    }    Matrix operator * (Matrix& b) {        Matrix c(n,b.m);        for(int i=1; i<=n; i++)            for(int j=1; j<=b.m; j++)                for(int k=1; k<=m; k++)                    c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mod;        return c;    }    void operator *= (Matrix& b) {        *this=*this*b;    }    Matrix operator ^ (LL b) {        Matrix ans(n,m,'e'),a=*this;        while(b>0) {            if(b&1)ans=ans*a;            a*=a;            b>>=1;        }        return ans;    }    void Output() {        for(int i=1; i<=n; i++) {            for(int j=1; j<=m; j++)                printf("%lld ",a[i][j]);            putchar('\n');        }    }};LL n;int main() {    Matrix a(1,2),b(2,2);    a[1][1]=a[1][2]=1;    b[1][1]=b[1][2]=b[2][1]=1;    while(scanf("%lld",&n)!=EOF) {        Matrix tmp=b^(n-2);        printf("%lld\n",(a*tmp)[1][1]);    }     return 0;}

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