SOJ-2984（Fibonacci，矩阵连乘简化计算）

来源：互联网发布：大数据发展规划编辑：程序博客网时间：2024/06/06 19:12
typedef int ll;#define MOD 10000int main(){//freopen("Fibonacci.in", "r", stdin);//freopen("Fibonacci.out", "w", stdout);    ll n;    ll t[10000];    while (scanf("%d", &n) == 1 && n != -1){        if (n == 0) {            printf("0\n");            continue ;        }        --n;//以下程序基于F0=1,F1=1，所以要做预处理        ll a = 1;        ll b = 1;        ll c = 1;        ll d = 0;        ll index = 0;        while (n) {            t[index++] = n % 2;            n /= 2;        }        ll i;        for (i = index - 2; i >= 0; --i) {            if (t[i] == 0) {                ll t1 = a, t2 = b, t3 = c, t4 = d;                a = t1 * t1 + t2 * t3;                b = t1 * t2 + t2 * t4;                c = t3 * t1 + t4 * t3;                d = t3 * t2 + t4 * t4;            } else {                ll t1 = a, t2 = b, t3 = c, t4 = d;                a = t1 * t1 + t2 * t3;                b = t1 * t2 + t2 * t4;                c = t3 * t1 + t4 * t3;                d = t3 * t2 + t4 * t4;a %= MOD;b %= MOD;c %= MOD;d %= MOD;                t1 = a, t2 = b, t3 = c, t4 = d;                a = t1 * 1 + t2 * 1;                b = t1 * 1 + t2 * 0;                c = t3 * 1 + t4 * 1;                d = t3 * 1 + t4 * 0;            }            a %= MOD;            b %= MOD;            c %= MOD;            d %= MOD;        }        printf("%d\n", a);    }    return 0;}