序列

问题描述 :

数列A满足An = An-1 + An-2 + An-3, n >= 3

编写程序，给定A0, A1 和 A2, 计算A99

输入:

输入包含多行数据

每行数据包含3个整数A0, A1, A2 (0 <= A0, A1, A2 <= 32767)

数据以EOF结束

输出:

对于输入的每一行输出A99的值

样例输入:

1 1 1

样例输出:

69087442470169316923566147

代码实现：

//* @author:
import java.util.*;
import java.math.BigInteger;
public class Main {

static String doAdd(String a, String b) { //两个大数相加的方法。
String str = "";
int lenA = a.length();
int lenB = b.length();
int maxLen = (lenA > lenB) ? lenA : lenB;
int minLen = (lenA < lenB) ? lenA : lenB;
String strTmp = "";
for (int i = maxLen - minLen; i > 0; i--) {
strTmp += "0";
}
// 把长度调整到相同
if (maxLen == lenA) {
b = strTmp + b;
} else
a = strTmp + a;
int JW = 0;// 进位
for (int i = maxLen - 1; i >= 0; i--) {
int tempA = Integer.parseInt(String.valueOf(a.charAt(i)));
int tempB = Integer.parseInt(String.valueOf(b.charAt(i)));

int temp;
if (tempA + tempB + JW >= 10 && i != 0) {
temp = tempA + tempB + JW - 10;
JW = 1;
} else {
temp = tempA + tempB + JW;
JW = 0;
}
str = String.valueOf(temp) + str;
}
return str;
}

public static void main(String[] args) {
Scanner in=new Scanner(System.in);

String a[]=new String [100];
while(in.hasNext()){
a[0]=Integer.toString(in.nextInt());
a[1]=Integer.toString(in.nextInt());
a[2]=Integer.toString(in.nextInt());
for(int i=3;i< 100;i++){
String temp=doAdd(a[i-1],a[i-2]);
a[i]=doAdd(temp,a[i-3]);
}

System.out.println(a[99]);
}
}
}

方法二：
import java.math.*;
import java.util.*;

public class Main
{
public static BigInteger calc(BigInteger a,BigInteger b,BigInteger c)
{
BigInteger now = c;
BigInteger last = b;
BigInteger llast = a;
BigInteger answer;
for(int i=0;i< 97;i++) {
answer = now.add(last);
answer = answer.add(llast);
llast = last;
last = now;
now = answer;
}
return now;
}

public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
while(in.hasNext()) {
int a0 = in.nextInt();
BigInteger A0 = BigInteger.valueOf(a0);
int a1 = in.nextInt();
BigInteger A1 = BigInteger.valueOf(a1);
int a2 = in.nextInt();
BigInteger A2 = BigInteger.valueOf(a2);
System.out.println(calc(A0,A1,A2));
}
}
}