JAVA高精度4_递推求值

第一道传送们：NYoj 113 某种序列

  一开始迭代顺序弄反了，a2=a3,a1=a2,a0=a1......太无脑了...因为从0开始，100-3，循环计算97次

import java.math.BigInteger;
import java.util.Scanner;

 class Main {
    public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            
            while(in.hasNext()) {           	
            BigInteger a0 = in.nextBigInteger();
            BigInteger a1 = in.nextBigInteger();
            BigInteger a2 = in.nextBigInteger();
            BigInteger a3 = BigInteger.ZERO;
            for(int i = 1; i<=97; i++) {
            	a3 = a0.add(a1).add(a2);
            	a0 = a1;
            	a1 = a2;
            	a2 = a3;
            		
            }
            System.out.println(a2);
			
            }
            
         }
    }

用数组套公式做也行：

import java.math.BigInteger;
import java.util.Scanner;

public class Main  
{  
    public static void main(String args[])  
    {  
        Scanner cin = new Scanner(System.in);     
        BigInteger a[] = new BigInteger[100];  
        while(cin.hasNext())  
        {  
            for(int i = 0; i <= 2; ++i)  
                a[i] = cin.nextBigInteger();  
            for(int i = 3; i <= 99; ++i)  
                a[i] = a[i - 1].add(a[i - 2]).add(a[i - 3]);  
            System.out.println(a[99]);  
        }  
    }  
}  

C/C++版：

也就是把数位存进数组里面，思路迭代和第一中Java一样：

#include <stdio.h>
#include <string.h>
#define MAX 1010
int a[MAX],b[MAX],c[MAX],sum[MAX];
void init(int x,int *a)
{
    int i=0;
    while(x)
    {
        a[i++]=x%10;
        x/=10;
    }
}
int main()
{
    int aa,bb,cc;
    while(scanf("%d %d %d",&aa,&bb,&cc)!=EOF)
    {
        int i,j,t;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        init(aa,a);init(bb,b);init(cc,c);
        /*if(aa==0&&bb==0&&cc==0)
        {
            printf("0
");
            continue;
        }*/
        //for(i=0;i<10;i++)
        //printf("%d %d %d ",a[i],b[i],c[i]);
        for(i=3;i<=99;i++)
        {
            memset(sum,0,sizeof(sum));
            for(j=0;j<MAX;j++)
            {
                sum[j]+=(a[j]+b[j]+c[j]);
                if(sum[j]>9)
                {
                    t=sum[j];
                    sum[j]=t%10;
                    sum[j+1]=t/10;
                }
            }
            for(j=0;j<MAX;j++)
            {
                a[j]=b[j];
                b[j]=c[j];
                c[j]=sum[j];
            }
        }
        for(j=MAX-1;j>0&&sum[j]==0;j--);
        for(;j>=0;j--)
        printf("%d",sum[j]);
        printf("
");
    }
    return 0;
}