[CQOI2013]二进制a+b

神做法见http://tieba.baidu.com/p/2354159387

DP，f[t][i][j][k][p]，表示枚举到第t位，A已放i个1，B已放j个1，C已放k个1，若p=0，表示该状态无进位；否则表示有进位。

 

/**

 * Problem:aplusb

 * Author:Shun Yao

 * Time:2013.5.27

 * Result:Accepted

 * Memo:DP

 */



#include <cstring>

#include <cstdlib>

#include <cstdio>



using namespace std;



long max(long x, long y) {

	return x > y ? x : y;

}

long min(long x, long y) {

	return x < y ? x : y;

}

void Modify(long &x, long y) {

	x = x == -1 ? y : min(x, y);

}



long get(long x){

	long s;

	for (s = 0; x; x >>= 1)

		if (1 & x)

			++s;

	return s;

}



long f[2][31][31][31][2];



int main() {

	long x, y, z, n = 0;

	freopen("aplusb.in", "r", stdin);

	freopen("aplusb.out", "w", stdout);

	scanf("%ld%ld%ld", &x, &y, &z);

	long a = get(x), b = get(y), c = get(z);

	x = max(x, max(y, z));

	for (; x; x >>= 1)

		++n;

	memset(f, -1, sizeof f);

	f[0][0][0][0][0] = 0;

	long cur = 0;

	for (long t = 0; t < n; ++t, cur ^= 1) {

		long aa = min(t, a), bb = min(t, b), cc = min(t, c);

		for (long i = 0; i <= aa; ++i)

			for (long j = 0; j <= bb; ++j)

				for (long k = 0; k <= cc; ++k) {

					if (f[cur][i][j][k][0] != -1) {

						long l = f[cur][i][j][k][0];

						Modify(f[cur ^ 1][i][j][k][0], l);

						if (i < a && k < c)

							Modify(f[cur ^ 1][i + 1][j][k + 1][0], l + (1 << t));

						if (i < a && j < b)

							Modify(f[cur ^ 1][i + 1][j + 1][k][1], l);

						if (j < b && k < c)

							Modify(f[cur ^ 1][i][j + 1][k + 1][0], l + (1 << t));

					}

					if(f[cur][i][j][k][1] != -1) {

						long l = f[cur][i][j][k][1];

						Modify(f[cur ^ 1][i][j][k + 1][0], l + (1 << t));

						if (i < a)

							Modify(f[cur ^ 1][i + 1][j][k][1], l);

						if (j < b)

							Modify(f[cur ^ 1][i][j + 1][k][1], l);

						if (i < a && j < b && k < c)

							Modify(f[cur ^ 1][i + 1][j + 1][k + 1][1], l + (1 << t));

                    }

				}

	}

	printf("%ld", f[cur][a][b][c][0]);

	fclose(stdin);

	fclose(stdout);

	return 0;

}