poj3070_斐波那契数列(Fibonacci)

用矩阵求斐波那契数列，快速幂log(n),只用求最后4位(加和乘的运算中前面的位数无用)

#include <stdio.h>
#include <stdlib.h>

int main()
{
    /*
    (x y)(x y)=  (x*x+y*s x*y+y*t) =(x*x+y*s  y*(x+t))
    (s t)(s t)   (s*x+t*s s*y+t*t)  (s*(x+t)  t*t+y*s)
    (x y)(a b)= (x*a+y*c x*b+y*d)
    (s t)(c d)  (s*a+t*c s*b+t*d)
    (a b)(f1=1)=(a+b)
    (c d)(f2=1) (c+d)
    */
    //(a,b)(c,d)一开始为E2
    long n,x,y,s,t,xx,yy,ss,tt,a,b,c,d,aa,bb,cc,dd;
    while (scanf("%ld",&n))
    {
        if (n==0)
        {
            printf("0
");
            continue;
        }
        else if (n==-1)
            break;
        n-=2;
        x=0;
        y=1;
        s=1;
        t=1;
        a=1;
        b=0;
        c=0;
        d=1;
        while (n)
        {
            if (n%2==1)
            {
                aa=a;
                bb=b;
                cc=c;
                dd=d;
                a=(x*aa+y*cc)%10000;
                b=(x*bb+y*dd)%10000;
                c=(s*aa+t*cc)%10000;
                d=(s*bb+t*dd)%10000;
            }
            xx=x;
            yy=y;
            ss=s;
            tt=t;
            x=(xx*xx+yy*ss)%10000;
            y=(yy*(xx+tt))%10000;
            s=(ss*(xx+tt))%10000;
            t=(tt*tt+yy*ss)%10000;
            n=n/2;
        }
        printf("%ld
",(c+d)%10000);
    }

    return 0;
}

#include <stdio.h>
#include <stdlib.h>

int main()
{
    /*
    (x y)(x y)=  (x*x+y*s x*y+y*t) =(x*x+y*s  y*(x+t))
    (s t)(s t)   (s*x+t*s s*y+t*t)  (s*(x+t)  t*t+y*s)
    (x y)(a b)= (x*a+y*c x*b+y*d)
    (s t)(c d)  (s*a+t*c s*b+t*d)
    (a b)(f1=1)=(a+b)
    (c d)(f2=1) (c+d)
    */
    //(a,b)(c,d)一开始为E2
    long n,x[32],y[32],s[32],t[32],a[32],b[32],c[32],d[32],w,r,i;
    x[0]=0;
    y[0]=1;
    s[0]=1;
    t[0]=1;
    a[0]=1;
    b[0]=0;
    c[0]=0;
    d[0]=1;
    for (i=1;i<32;i++)
    {
        x[i]=(x[i-1]*x[i-1]+y[i-1]*s[i-1])%10000;
        y[i]=(y[i-1]*(x[i-1]+t[i-1]))%10000;
        s[i]=(s[i-1]*(x[i-1]+t[i-1]))%10000;
        t[i]=(t[i-1]*t[i-1]+y[i-1]*s[i-1])%10000;
    }
    while (scanf("%ld",&n))
    {
        if (n==0)
        {
            printf("0
");
            continue;
        }
        else if (n==-1)
            break;
        n-=2;
        w=0;
        r=0;
        while (n)
        {
            if (n%2==1)
            {
                a[r+1]=(x[w]*a[r]+y[w]*c[r])%10000;
                b[r+1]=(x[w]*b[r]+y[w]*d[r])%10000;
                c[r+1]=(s[w]*a[r]+t[w]*c[r])%10000;
                d[r+1]=(s[w]*b[r]+t[w]*d[r])%10000;
                r++;
            }
            w++;
            n=n/2;
        }
        printf("%ld
",(c[r]+d[r])%10000);
    }
    return 0;
}