一道算法题，求最优解法？

hot_ice

import java.io.*;
import java.sql.*;
import java.math.BigInteger;
import java.math.*;

public class jie1
{
    public static void main(String[] args)
    {
    int c=400;
    int i=0;
    long t=System.currentTimeMillis();

    BigInteger a = new BigInteger("0");
    BigInteger b = new BigInteger("1");
    BigInteger d = new BigInteger("1");
    BigInteger ten =new BigInteger("10");
    while(d.compareTo(ten.pow(c))<0){i++;
    BigInteger temp =d.add(a.add(b)).multiply(new BigInteger("2"));
    BigInteger tempa = temp.subtract( b);
    BigInteger tempb = temp.subtract( a);
    BigInteger tempc = temp.add( d);
    a = tempa;
    b = tempb;
    d = tempc;
    System.out.println(i+":"+a.toString()+" "+b.toString()+" "+d.toString());
   }
   i=(int)(System.currentTimeMillis()-t);
   System.out.println("用时"+i+"毫秒");
    }
}

yuanchengjun

我还没编译成Release的。n == 2000，用时不到1毫秒。
浮点在判断相等的时候，肯定会有问题的。

#include <time.h> #include <stdio.h> int main(int argc, char* argv[]) { &nbsp;&nbsp;&nbsp;&nbsp;int a; &nbsp;&nbsp;&nbsp;&nbsp;int n = 2000; &nbsp;&nbsp;&nbsp;&nbsp;int cnt = 0; &nbsp;&nbsp;&nbsp;&nbsp;int r [2000]; &nbsp;&nbsp;&nbsp;&nbsp;int i; &nbsp;&nbsp;&nbsp;&nbsp;int AMax = n - 2; &nbsp;&nbsp;&nbsp;&nbsp;int CxCMax = (n - 1) * (n - 1); &nbsp;&nbsp;&nbsp;&nbsp;clock_t begin, end; &nbsp;&nbsp;&nbsp;&nbsp;begin = clock(); &nbsp;&nbsp;&nbsp;&nbsp;for (a = 2; a <= AMax; a ++) &nbsp;&nbsp;&nbsp;&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (a * a + (a + 1) * (a + 1) <= CxCMax) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;r[cnt ++] = a; &nbsp;&nbsp;&nbsp;&nbsp;} &nbsp;&nbsp;&nbsp;&nbsp;end = clock(); &nbsp;&nbsp;&nbsp;&nbsp;for (i = 0; i < cnt; i++) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;printf("%d^2 + %d^2 = %d\n", r[i], r[i] + 1, r[i] * r[i]); &nbsp;&nbsp;&nbsp;&nbsp;printf(" -> time %ld, total %d\n", end - begin, cnt); &nbsp;&nbsp;&nbsp;&nbsp;return 0; }

1387^2 + 1388^2 = 1923769
1388^2 + 1389^2 = 1926544
1389^2 + 1390^2 = 1929321
1390^2 + 1391^2 = 1932100
1391^2 + 1392^2 = 1934881
1392^2 + 1393^2 = 1937664
1393^2 + 1394^2 = 1940449
1394^2 + 1395^2 = 1943236
1395^2 + 1396^2 = 1946025
1396^2 + 1397^2 = 1948816
1397^2 + 1398^2 = 1951609
1398^2 + 1399^2 = 1954404
1399^2 + 1400^2 = 1957201
1400^2 + 1401^2 = 1960000
1401^2 + 1402^2 = 1962801
1402^2 + 1403^2 = 1965604
1403^2 + 1404^2 = 1968409
1404^2 + 1405^2 = 1971216
1405^2 + 1406^2 = 1974025
1406^2 + 1407^2 = 1976836
1407^2 + 1408^2 = 1979649
1408^2 + 1409^2 = 1982464
1409^2 + 1410^2 = 1985281
1410^2 + 1411^2 = 1988100
1411^2 + 1412^2 = 1990921
1412^2 + 1413^2 = 1993744
1413^2 + 1414^2 = 1996569
-> time 0, total 1412

hot_ice

原帖由 yuanchengjun 于 2008-4-26 08:37 发表
我还没编译成Release的。n == 2000，用时不到1毫秒。
浮点在判断相等的时候，肯定会有问题的。

#include
#include

int main(int argc, char* argv[])
{
&nbsp;&nbsp;&nbsp;&nbsp;int a;
&nbsp;&nb ...

貌似写的不对