In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
099999999991000000000-1Sample Output
0346266875Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
题目：一个矩阵快速幂就可以
代码示例：
#define ll long longconst ll maxn = 1e6+5;const ll mod = 1e4;const double eps = 1e-9;const double pi = acos(-1.0);const ll inf = 0x3f3f3f3f;ll n;struct mat{ ll a[2][2];};mat mul(mat A, mat B){ mat r; memset(r.a, 0, sizeof(r.a)); for(ll i = 0; i < 2; i++){ for(ll j = 0; j < 2; j++){ for(ll k = 0; k < 2; k++){ r.a[i][j] += (A.a[i][k]*B.a[k][j])%mod; r.a[i][j] %= mod; } } } return r;}mat qpow(mat A, ll x){ mat B; B.a[0][0] = B.a[1][1] = 1; // 单位矩阵 B.a[0][1] = B.a[1][0] = 0; while(x){ if (x&1) B = mul(B, A); A = mul(A, A); x >>= 1; } return B;}int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); while(~scanf("%lld", &n)){ if (n == -1) break; mat a; a.a[0][0] = a.a[0][1] = a.a[1][0] = 1; a.a[1][1] = 0; if (n == 0) printf("0\n"); else if (n == 1) printf("1\n"); else { a = qpow(a, n-1); printf("%d\n", a.a[0][0]%mod); } } return 0;}