JZOJ.1240. Fibonacci sequence

Problem

Solution

矩阵乘法。设有一个矩阵a为[sum,x,y,z]，其中sum表示Σni=1F[i]x,y表示F[i−2],F[i−1]，z表示F[i]。那么通过一个4*4的矩阵b

[1,0,0,0]

[0,0,0,0]

[1,1,0,1]

[1,0,1,1]

相乘变成a[sum’,y,z,new] （显而易见，new=y+z，sum’=sum+y+z）

具体怎么矩阵乘法呢？

首先我们先求出Σyi=1F[i]再求出Σx−1i=1F[i]两者相减即为答案。

我们可以想到矩阵快速幂，将矩阵b变成(矩阵b)y。然后与a相乘。

求(矩阵b)x−1也是如此。

不要忘了%10000。

Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#define LL long long
#define fo(i,a,b) for(i=a;i<=b;i++)
#define mo 10000
using namespace std;
const LL b[5][5]={{0,0,0,0,0},{0,1,0,0,0},{0,0,0,0,0},{0,1,1,0,1},{0,1,0,1,1}};
const LL h[5][5]={{0,0,0,0,0},{0,1,0,0,0},{0,0,1,0,0},{0,0,0,1,0},{0,0,0,0,1}};
const LL a[5]={0,4,1,1,2};
LL e[5][5],c[5],ans,A1,A2;
int i,j,k,l,n,m,T,s,x,y,p;
LL jzcf(int x)
{
int i,j,k,p=x;
LL d[5][5],f[5][5],g[5][5],c[5],A1;
memcpy(g,b,sizeof(b));
memcpy(d,h,sizeof(h));
while (p>0)
{
if (p%2==1)
{
memset(f,0,sizeof(f));
fo(k,1,4)
fo(i,1,4)
fo(j,1,4)
f[i][j]=(f[i][j]+d[i][k]*g[k][j])%mo;
memcpy(d,f,sizeof(f));
}
memset(f,0,sizeof(f));
fo(k,1,4)
fo(i,1,4)
fo(j,1,4)
f[i][j]=(f[i][j]+g[i][k]*g[k][j])%mo;
memcpy(g,f,sizeof(f));
p=p/2;
}
A1=0;
memset(c,0,sizeof(c));
fo(k,1,4)
fo(j,1,4)
c[j]=(c[j]+a[k]*d[k][j])%mo;
A1=c[1];
return A1;
}
LL res(int x)
{
LL rs;
if (x==0) return 0;
if (x==1) return 1;
if (x==2) return 2;
if (x==3) return 4;
rs=jzcf(x-3);
return rs;
}
int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%d%d",&x,&y);
A1=res(y);
A2=res(x-1);
ans=(A1-A2+mo)%mo;
printf("%lld\n",ans);
}
}

——2016.7.12