csu 1104 Fibonacci Numbers

超过8位只输出高低四位，中间用...隔开：

Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”). 

高四位的输出：利用通项公式，先求出取对数的小数部分，然后取前四位即可。

第四位：直接递推会严重超时！！所以只能用矩阵幂的方法（其实题目提示了用线性代数知识：Use your linear algebra knowledge）

/* csu 1104 */
# include <stdio.h>
# include <math.h>
# include <time.h>
# define MODN 10000

typedef struct Matrix
{
    int a,b,c,d;
} mat;


int f[] = {0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,
           2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,
           317811,514229,832040,1346269,2178309,3524578,5702887,9227465,
           14930352,24157817,39088169,63245986
          };


mat cur, mpow, tmp;

int last_4_dig(int n)
{
    int i, k, m;

    m = (n-1)>>1;
    cur.a = cur.d = 1, cur.b = cur.c = 0;
    mpow.a = mpow.b = mpow.c = 1, mpow.d = 2;

    for (k = 31; k >= 0; --k)
        if ((m>>k) & 0x1) break;
    for (i = 0; i <= k; ++i)
    {
        if ((m>>i) & 0x1)
        {
            tmp = cur;
            cur.a = (tmp.a*mpow.a+tmp.b*mpow.c) % MODN;
            cur.b = (tmp.a*mpow.b+tmp.b*mpow.d) % MODN;
            cur.c = (tmp.c*mpow.a+tmp.d*mpow.c) % MODN;
            cur.d = (tmp.c*mpow.b+tmp.d*mpow.d) % MODN;
        }
        tmp = mpow;
        mpow.a = (tmp.a*tmp.a + tmp.b*tmp.c) % MODN;
        mpow.b = (tmp.b * (tmp.a+tmp.d)) % MODN;
        mpow.c = (tmp.c * (tmp.a+tmp.d)) % MODN;
        mpow.d = (tmp.c*tmp.b + tmp.d*tmp.d) % MODN;
    }
    if (n & 0x1) return (cur.a+cur.b)%MODN;
    else return (cur.c+cur.d)%MODN;
}

//int last_4_dig(int n)　　// 严重 TLE !!
//{
//    int a, b, i;
//
//    i = 38;
//    a = f[38] % 10000;
//    b = f[39] % 10000;
//
//    while (i+1 < n)
//    {
//        a = (a + b) % 10000;
//        b = (a + b) % 10000;
//        i += 2;
//    }
//    return (n&0x1) ? b:a;
//}

int main()
{
    int n;
    double first_4_dig;

    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);

    while (~scanf("%d", &n))
    {
        if (n < 40)
        {
            printf("%d\n", f[n]);
            continue;
        }
        else
        {
            first_4_dig = n*log10(0.5*(1+sqrt(5)))-0.5*log10(5.0);
            first_4_dig -= (int)first_4_dig;
            first_4_dig = pow(10, first_4_dig);
            while (first_4_dig < 1000) first_4_dig *= 10;
            printf("%d", (int)first_4_dig);
            printf("...%04d\n", last_4_dig(n));
        }
    }

    printf("time cost %.3lf\n", (double)clock()/CLOCKS_PER_SEC);

    return 0;
}

其实原来弄错了，以为是从第1项开始依次是0、1、1，中间还怀疑示例数据错了。。。o(╯□╰)o

所以在求后四位数那里，可以稍微改一下，返回时直接返回矩阵的某一个元素即可（f[0]=0,f[1]=1）