Codeforces Round #258 (Div. 2) B. Jzzhu and Sequences（矩阵快速幂）_html/css

Codeforces Round #258 (Div. 2) B. Jzzhu and Sequences（矩阵快速幂）_html/css_WEB-ITnose

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发布： 2016-06-24 12:01:33

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题目链接：http://codeforces.com/problemset/problem/450/b

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B. Jzzhu and Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Jzzhu has invented a kind of sequences, they meet the following property:

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You are given x and y, please calculate fn modulo 1000000007 (109?+?7).

Input

The first line contains two integers x and y (|x|,?|y|?≤?109). The second line contains a single integer n (1?≤?n?≤?2·109).

Output

Output a single integer representing fn modulo 1000000007 (109?+?7).

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Sample test(s)

input

2 33

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output

input

0 -12

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output

1000000006

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Note

In the first sample, f2?=?f1?+?f3, 3?=?2?+?f3, f3?=?1.

In the second sample, f2?=??-?1; ?-?1 modulo (109?+?7) equals (109?+?6).

代码如下：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;struct A{    int mat[2][2];};A d,f;__int64 n,mod;A mul(A a,A b){    A t;    memset(t.mat,0,sizeof(t.mat));    for(int i=0;i<n;i++)    {        for(int k=0;k<n;k++)        {            if(a.mat[i][k])                for(int j=0;j<n;j++)                {                    t.mat[i][j]+=a.mat[i][k]*b.mat[k][j];                    t.mat[i][j]%=mod;                }        }    }    return t;}A quickP(int k){    A p = d ,m;    memset(m.mat,0,sizeof(m.mat));    for(int i=0;i<n;++i)//单位矩阵    {        m.mat[i][i]=1;    }    while(k)    {        if(k & 1)            m=mul(m,p);        p=mul(p,p);        k >>= 1 ;    }    return m;}int main(){    n=2;    int k,t;__int64 x,y,z;    while(scanf("%I64d%I64d",&x,&y)!=EOF)    {        int s=0;        scanf("%I64d",&z);        mod=1000000007;        if(z == 1)        {            if(x < 0)                printf("%I64d\n",x+mod);            else                printf("%I64d\n",x);            continue;        }        d.mat[0][1]=-1;d.mat[1][1] = 0;        d.mat[0][0]=d.mat[1][0]=1;        A ret=quickP(z-2);//z-2 乘的次数        __int64 ans=(ret.mat[0][0]*y%mod+ret.mat[0][1]*x%mod)%mod;        if(ans < 0)            ans+=mod;        printf("%I64d\n",ans);    }    return 0;}

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