[luoguP1489] 猫狗大战（DP）

传送门

类似背包的做法。

f[i][j]表示是否能放i个物品，价格为j

#include <cstdio>
#include <iostream>
#define N 8001

int n, sum;
int a[201], f[201][N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

int main()
{
	int i, j, k;
	n = read();
	for(i = 1; i <= n; i++)
	{
		a[i] = read();
		sum += a[i];
	}
	f[0][0] = 1;
	for(i = 1; i <= n; i++)
		for(j = n >> 1; j >= 1; j--)
			for(k = sum >> 1; k >= a[i]; k--)
				f[j][k] = f[j][k] | f[j - 1][k - a[i]];
	for(i = j = sum >> 1; ; i--, j++)
	{
		if(f[n >> 1][i])
		{
			printf("%d %d\n", i, sum - i);
			return 0;
		}
		if(f[n >> 1][j])
		{
			printf("%d %d\n", sum - j, j);
			return 0;
		}
	}
	return 0;
}