HITOJ 2060 类似斐波那契数列（一段和取模）

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http://acm.hit.edu.cn/hoj/problem/view?id=2060

As we know , the Fibonacci numbers are defined as follows:

""""

Given two numbers a and b , calculate . """"

Input

The input contains several test cases. Each test case consists of two non-negative integer numbers a and b (0 ≤ a ≤ b ≤1,000,000,000). Input is terminated by a = b = 0.

Output

For each test case, output S mod 1,000,000,000, since S may be quite large.

Sample Input

1 13 510 10000 0

Sample Output

116496035733

题目大意：给一个类似于Fibonacci的数列，求第a项到第b项的和取模。

解题思路：构造一个3*3的矩阵，利用矩阵连乘的思想求解

/*This Code is Submitted by life4711 for Problem 2060 at 2014-07-25 14:52:17*/#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>using namespace std;typedef long long LL;const int N=3;const LL MOD=1000000000;struct Matrix{    LL m[N][N];};Matrix I={    1,0,0,    0,1,0,    0,0,1};Matrix multi(Matrix a,Matrix b){    Matrix c;    for(int i=0; i<N; i++)        for(int j=0; j<N; j++)        {            c.m[i][j]=0;            for(int k=0; k<N; k++)            {                c.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;            }            c.m[i][j]=c.m[i][j]%MOD;        }    return c;}Matrix quick_mod(Matrix a,LL k){    Matrix ans=I;    while(k!=0)    {        if(k&1)        {            ans=multi(ans,a);        }        k>>=1;        a=multi(a,a);    }    return ans;}int main(){    LL n,m;    while(~scanf("%lld%lld",&n,&m))    {        if(n==0&&m==0)            break;        Matrix A={1,1,1,                  0,1,1,                  0,1,0};        if(n==0)        {            Matrix x1=quick_mod(A,m-1);            LL s=(x1.m[0][0]*2%MOD+x1.m[0][1]+x1.m[0][2])%MOD;            if(s<0)                s+=MOD;            printf("%lld\n",s);        }        else if(n==1)        {            Matrix x1=quick_mod(A,m-1);            LL s=(x1.m[0][0]*2%MOD+x1.m[0][1]+x1.m[0][2])%MOD;            s=(s-1)%MOD;            if(s<0)                s+=MOD;            printf("%lld\n",s);        }        else        {            Matrix x1=quick_mod(A,m-1);            LL s1=(x1.m[0][0]*2%MOD+x1.m[0][1]+x1.m[0][2])%MOD;            x1=quick_mod(A,n-2);            LL s2=(x1.m[0][0]*2%MOD+x1.m[0][1]+x1.m[0][2])%MOD;            s1=(s1%MOD-s2%MOD)%MOD;            if(s1<0)                s1+=MOD;            printf("%lld\n",s1);        }    }    return 0;}

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