project euler 120

Square remainders

Let r be the remainder when (a?1)n + (a+1)n is divided by a2.

For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 ≡ 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.

For 3 ≤ a ≤ 1000, find ∑rmax.

---

平方余数

记r是(a?1)n + (a+1)n被a2除所得的余数。

例如，如果a = 7而n = 3，则r = 42：63 + 83 = 728 ≡ 42 mod 49。随着n的变化，r也会随之变化，但是对a = 7，可以得出rmax = 42。

对于3 ≤ a ≤ 1000，求∑rmax。

package projecteuler;

import junit.framework.TestCase;

public class Prj120 extends TestCase {

	public void testSquareRemainders() {

		int sum = 0;
		for (int a = 3; a <= 1000; a++) {
			int rmax = getRmax(a);
			sum += rmax;
		}
		System.out.println("sum=" + sum);
	}

	/**
	 * na[1 + (-1)^(n-1)] + [1 + (-1)^n];
	 * 
	 * @param a
	 * @return
	 */
	int getRmax(int a) {

		int nid = 2;
		int rmax = 2;
		for (int n = 1; n <= 2*a; n = n + 2) {
			int r = n * a * 2;
			r %= (a * a);
			if (r > rmax) {
				rmax = r;
				nid = n;
			}
		}
		String fstr = "n=%d,a=%d,rmax=%d";
		fstr = String.format(fstr, nid, a, rmax);
		System.out.println(fstr);

		return rmax;
	}
}