Tribonacci数列是斐波那挈数列的扩展
很有趣的，我们可以发现 
这是Tribonacci数列的一些深入研究
下面是贴代码的时间了：
解法一（半产品）
这种方法就不解释了，不懂就去看看最笨的方法递归求解，而这是对递归求解的优化
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
long l = scanner.nextLong();
long r = scanner.nextLong();
long sum = 0;
sum = sum_tribonacci(l, r);
System.out.println(sum % 1000000007);
// for (long i = 0; i <= 100l; i++) {
// sum = sum_tribonacci(0, i);
// System.out.println(i + "---------------------------" + sum
// % 1000000007l);
// }
}
}
public static long sum_tribonacci(long l, long r) {
long n1 = 0, n2 = 0, n3 = 0;
long u1, u2;
long sum = 0;
if (r < 3) {
return (r - l + 1);
}
n1 = 0;
n2 = 1;
n3 = 2;
// n1 = 1;
// n2 = 1;
// n3 = 1;
long n4 = 0;
if (l < 3) {
// for (long i = 3; i <= r; i++) {
for (long i = 3; i <= r + 1; i++) {
n4 = n3 + n2 + n1;
n1 = n2;
n2 = n3;
n3 = n4;
}
sum = n3 - l;
return sum;
} else {
// for (long i = 3; i <= l; i++) {
for (long i = 3; i <= l; i++) {
n4 = n3 + n2 + n1;
n1 = n2;
n2 = n3;
n3 = n4;
}
long sum1 = n3;
// sum+=n4;
for (long i = l + 1; i <= r + 1; i++) {
n4 = n3 + n2 + n1;
n1 = n2;
n2 = n3;
n3 = n4;
}
long sum2 = n3;
return sum2 - sum1;
}
}
}
解法二 
所以这里关键是要求出矩阵A
可以用前5项求出矩阵（数列{1、2、3、6、11…}依题意这是数列{1、1、1、3、5…}前n-1项的和，依然满足Tribonacci规则）
这里矩阵的n次幂，我采用的是二分法。
1 import java.math.BigDecimal;
2 import java.util.Scanner;
3
4 public class CopyOfMain2 {
5 public static BigDecimal i1000000007 = new BigDecimal(
6 String.valueOf(1000000007));
7 //这里定义了四个数，其实是为了下面BigDecimal数组的初始化做准备
8 public static BigDecimal i3 = new BigDecimal(String.valueOf(3));
9 public static BigDecimal i2 = new BigDecimal(String.valueOf(2));
10 public static BigDecimal i1 = new BigDecimal(String.valueOf(1));
11 public static BigDecimal i0 = new BigDecimal(String.valueOf(0));
12
13 public static void main(String[] args) {
14 Scanner scanner = new Scanner(System.in);
15
16 while (scanner.hasNext()) {
17
18 long l = scanner.nextLong();
19 long r = scanner.nextLong();
20 BigDecimal[] sum = {i0,i0};
21
22
23 sum = Tribonacci(l, r).divideAndRemainder(i1000000007);
24 if(sum[1].longValue()<0)System.out.println(sum[1].add(i1000000007));
25 else System.out.println(sum[1]);
26 //这是比较笨的方法，用这种遍历的求和方式当然会超时
27 // for (long i = l; i <= r; i++) {
28 // sum += Tribonacci(i);
29 // }
30 // for(long i =0; i <= 100l; i++){
31 // sum=Tribonacci(0,i);
32 // System.out.println(i+"---------------------------"+sum %
33 // 1000000007l);
34 // }
35 }
36 }
37
38 public static BigDecimal Tribonacci(long l, long r) {
39 BigDecimal sum = i0;
40 if (r < 3) {
41 for (long i = l; i <= r; i++) {
42 sum = new BigDecimal(String.valueOf(r - l + 1));
43 }
44 return sum;
45 }
46 BigDecimal[][] base = { { i1, i1, i0 }, { i1, i0, i1 }, { i1, i0, i0 } };
47 if (l >= 3l) {
48 BigDecimal[][] res1 = matrixPower(base, l - 3);
49
50 BigDecimal[][] res = matrixPower(base, r - 2);
51 // long[][] res = muliMatrix(res1,matrixPower(base, r- l+1));
52 return (res[0][0].subtract(res1[0][0])).multiply(i3)
53 .add((res[1][0].subtract(res1[1][0])).multiply(i2))
54 .add((res[2][0].subtract(res1[2][0])));
55 } else {
56 BigDecimal[][] res = matrixPower(base, r - 2);
57 return (res[0][0].multiply(i3).add(res[1][0].multiply(i2))
58 .add(res[2][0]).subtract(new BigDecimal(String.valueOf(l))));
59 }
60
61 }
62 //求Tribonacci数列每一项的方法
63 // public static long Tribonacci(long n) {
64 // if (n == 0l || n == 1l || n == 2l) {
65 // return 1;
66 // } else if (n == 3l)
67 // return 3;
68 // long[][] base = { { 1l, 1l, 0l}, { 1l, 0l, 1l }, { 1l, 0l, 0l } };
69 // long sum = 0l;
70 //
71 // long[][] res = matrixPower(base, n - 3l);
72 //
73 // return 3l * res[0][0] + res[1][0] + res[2][0];
74 // }
75
76 public static BigDecimal[][] matrixPower(BigDecimal[][] base, long p) {
77 BigDecimal[][] res = new BigDecimal[base.length][base[0].length];
78 for (int i = 0; i < res.length; i++) {
79 for (int j = 0; j< res[0].length; j++) {
80 if (i == j) {
81 res[i][j] = i1;
82 } else
83 res[i][j] = i0;
84 }
85
86 }
87 BigDecimal tmp[][] = base;
88 for (; p != 0; p >>= 1) {
89 if ((p & 1) != 0) {
90 res = muliMatrix(res, tmp);
91 }
92 tmp = muliMatrix(tmp, tmp);
93 }
94 return res;
95 }
96
97 private static BigDecimal[][] muliMatrix(BigDecimal[][] m1,
98 BigDecimal[][] m2) {
99 BigDecimal[][] res = new BigDecimal[m1.length][m2[0].length];
100 for (int i = 0; i < res.length; i++) {
101 for (int j = 0; j< res[0].length; j++) {
102 res[i][j] = i0;
103 }
104 }
105 for (int i = 0; i < m1.length; i++) {
106 for (int j = 0; j < m2[0].length; j++) {
107 for (int k = 0; k < m2.length; k++) {
108
109 res[i][j] = res[i][j].add(m1[i][k].multiply(m2[k][j])).divideAndRemainder(i1000000007)[1];
110
111 }
112 }
113 }
114 return res;
115
116 }
117
118 }
时间： 2024-10-12 20:45:18