M斐波那契數列 hdu 4549
阿新 • • 發佈:2019-02-13
M斐波那契數列
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 520 Accepted Submission(s): 146
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
現在給出a, b, n,你能求出F[n]的值嗎? Input 輸入包含多組測試資料;
每組資料佔一行,包含3個整數a, b, n( 0 <= a, b, n <= 10^9 ) Output 對每組測試資料請輸出一個整數F[n],由於F[n]可能很大,你只需輸出F[n]對1000000007取模後的值即可,每組資料輸出一行。 Sample Input 0 1 0 6 10 2 Sample Output 0 60
package hpu; import java.util.Scanner; public class HDU4579_3 { public static int a, b, n; public static long mod = 1000000007; public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNextInt()) { a = sc.nextInt(); b = sc.nextInt(); n = sc.nextInt(); if(n == 0) { System.out.println(a); }else if(n == 1) { System.out.println(b); } else { sovle(); } } } private static void sovle() { long[][] result = {{1, 0}, {0, 1}}; long[][] f = {{1, 1},{1, 0}}; while(n > 0) { if((n&1) == 1) { matrixMul(result, f); } matrixMul(f, f); n >>= 1; } long bn = result[0][1]; long an = result[1][1]; long bbn = powMod(b, bn, mod); long aan = powMod(a, an, mod); long r = bbn*aan%mod; System.out.println(r); } //矩陣相乘 private static void matrixMul(long[][] a, long[][] b) { long c[][] = new long[2][2]; for(int i=0; i<2; i++) { for(int j=0; j<2; j++) { for(int k=0; k<2; k++) { c[i][j] = (c[i][j]+a[i][k]*b[k][j])%(mod-1);//%(mod-1)的原因:費馬小定理(a^(m-1) = 1%m)(m==素數)) } } } for(int i=0; i<2; i++) { for(int j=0; j<2; j++) { a[i][j] = c[i][j]; } } } //求a^n%m private static long powMod(long a, long n, long m) { long c = 1; while(n > 0) { if((n&1) == 1) { c = c*a%m; } a = a*a%m; n >>= 1; } return c; } //求a^n%m(遞迴型) // private static long powMod(long a, long n, long m) { // if(n==0) return 1; // long x = powMod(a, n/2, m); // long ans = (long)x*x%m; // if(n%2 == 1) ans = ans*a%m; // return ans; // } }