不多解释，详情参阅我的相关文章。
Add()的两个参数都是逆序的整数数组，结果存在num1中。
以下是本题代码：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define LEN 1000
int Max(int a, int b)
{
if(a > b)
return a;
return b;
}
void Add(char *num1, char *num2)
{
int len1, len2;
int i, j;
i = LEN;
while(num1[--i] == 0 && i > 0);
len1 = i;
i = LEN;
while(num2[--i] == 0 && i > 0);
len2 = i;
int lenm = Max(len1, len2);
for(i = 0; i <= lenm; i++)//add
num1[i] += num2[i];
int t = 0;
int t2;
for(i = 0; i <= lenm + 2; i++)//carry bit
{
t2 = num1[i] + t;
num1[i] = t2 % 10;
t = t2 / 10;
}
}
void ToAr(char *A, int n)
{
int i, j;
i = 0;
while(n)
{
A[i++] = n % 10;
n /= 10;
}
}
void Copy(char *t, char *f)
{
for(int i = 0; i < LEN; i++)
t[i] = f[i];
}
int main()
{
int i, j;
char A0[LEN], A1[LEN], A2[LEN], A3[LEN];
char str[LEN];
while(gets(str) != NULL)
{
memset(A0, 0, sizeof(A0));
memset(A1, 0, sizeof(A0));
memset(A2, 0, sizeof(A0));
memset(A3, 0, sizeof(A0));
int a0, a1, a2;
sscanf(str, "%d%d%d", &a0, &a1, &a2);
ToAr(A0, a0);
ToAr(A1, a1);
ToAr(A2, a2);
for(j = 1; j <= 97; j++)
{
Add(A0, A1);
Add(A0, A2);
Copy(A3, A0);
Copy(A0, A1);
Copy(A1, A2);
Copy(A2, A3);
}
i = LEN;
while(A3[--i] == 0 && i > 0);
if(i == 0)
printf("0\n");
else
{
for(; i >= 0; i--)
printf("%d", A3[i]);
putchar(10);
}
}
//system("pause");
}
下面是java代码：
import java.math.*;
import java.util.*;
public class Main 
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
try
{
while(true)
{
BigInteger b1, b2, b3, b4;
b1 = sc.nextBigInteger();
b2 = sc.nextBigInteger();
b3 = sc.nextBigInteger();
b4 = new BigInteger("0");
for(int i = 0; i <= 96; i++)
{
b1 = b1.add(b2);
b1 = b1.add(b3);
b4 = b1;
b1 = b2;
b2 = b3;
b3 = b4;
}
System.out.println(b4);
}
}catch(Exception e){};
}
}
posted on 2012-08-11 09:53
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