标签：and NPU des sso inf nta ons rod enc
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).
0 9 999999999 1000000000 -1Sample Output
0 34 626 6875Hint
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 long
const 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;
}
标签：and NPU des sso inf nta ons rod enc
原文地址：https://www.cnblogs.com/ccut-ry/p/8973153.html
