需求是这样的：
有44个数字A1-A44：1420, 1198, 493, 300, 3006, 881, 2450, 1372, 3643, 1350, 598, 615, 740, 820, 841, 820, 1025, 885, 270, 270, 270, 270, 270, 1735, 885, 4719, 1661, 1435, 2255, 5895, 6465, 870, 330, 750, 1365, 2085, 3370, 1995, 7990, 381, 2048, 840, 6880, 7723

另有3个数字S1-S3：31827, 37944, 15713

44个数据分成三组，使每一组的和分别等于S1、S2、S3，即第一组的和等于S1，第二组的和等于S2，第三组的和等于S3.。
44个数据中的每一个数，要么属于第一组，要么属于第二组，要么属于第三组，不能即属于第一组又属于第二组。

结果：37944等于1420,1198,493,300,3006,881,2450,1372,3643,1350,598,615,740,820,841,820,1025,885,270,270,270,270,270,1735,885,4719,1435,870,330,750,1365,2048,
31827等于
1661,2255,5895,6465,2085,3370,1995,381,840,6880,
15713等于
剩下的

public class test {
	static int sum = 0;
	static int sum2 = 0;
	static int c = 0;
	static int c2 = 0;
	static int n = 0;
	static int[] arr;
	static int[] temp;
	static int[] temp2;

	static int t = 0;
	static int t2 = 0;

	public static void main(String[] args) {
		n = 44;
		c = 37944;
		c2 = 31827;
		temp = new int[n];
		temp2 = new int[n];

		arr = new int[] { 1420, 1198, 493, 300, 3006, 881, 2450, 1372, 3643, 1350, 598, 615, 740, 820, 841, 820, 1025,
				885, 270, 270, 270, 270, 270, 1735, 885, 4719, 1661, 1435, 2255, 5895, 6465, 870, 330, 750, 1365, 2085,
				3370, 1995, 7990, 381, 2048, 840, 6880, 7723 };
		search(0);
		if (t == 0) {
			System.out.println("No Solution!");
		}
	}

	public static void search(int m) {
		if (sum == c) {
			t = 1;
			int[] sub = new int[m];
			for (int i = 0; i < m; i++) {
				if (temp[i] == 1) {
					sub[i] = arr[i];
				}
			}
			int[] bb = buji(sub);
			search2(bb, 0);
			for (int i = 0; i < sub.length; i++) {
				if (sub[i] == 0)
					continue;
				System.out.print(sub[i] + ",");
			}
			return;
		}
		if (sum > c || m == n) {
			return;
		}
		if (sum + arr[m] <= c) {
			sum += arr[m];
			temp[m] = 1;
			search(m + 1);
			temp[m] = 0;
			sum -= arr[m];
		}
		search(m + 1);
	}

	public static void search2(int[] bb, int m2) {
		int n = bb.length;
		if (sum2 == c2) {
			t2 = 1;
			System.out.println("");
			System.out.println("");
			for (int i = 0; i < m2; i++) {
				if (temp2[i] == 1) {
					System.out.print(bb[i] + ",");
				}
			}
			System.out.println("");
			System.exit(0);
			return;
		}
		if (sum2 > c2 || m2 == n) {
			return;
		}
		if (sum2 + bb[m2] <= c2) {
			sum2 += bb[m2];
			temp2[m2] = 1;
			search2(bb, m2 + 1);
			temp2[m2] = 0;
			sum2 -= bb[m2];
		}
		search2(bb, m2 + 1);
	}

	// 从求一个集合和子集的补集
	public static int[] buji(int[] b) {
		int[] arr2 = new int[n];
		arr2 = arr;
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < arr2.length; j++) {
				if (b[i] == arr2[j]) {
					arr2[j] = 0;
					break;
				}
			}
		}
		int k = arr.length - b.length;
		int[] cc = new int[k];
		for (int i = 0, j = 0; i < arr2.length; i++) {
			if (arr2[i] == 0) {
				continue;
			}
			if (j < k) {
				cc[j] = arr2[i];
				j++;
			}
		}
		return cc;

	}
}