Atcoder Tenka1 Programmer Contest 2019 D Three Colors

题意：
有(n)个石头，每个石头有权值，可以给它们染'R', 'G', 'B'三种颜色，如下定义一种染色方案为合法方案：

• 所有石头都染上了一种颜色
• 令(R, G, B)为染了'R', 染了'G', 染了'B'的所有石头的权值和，存在一个三角形的三边为(R, G, B)

求合法方案数模(998244353)

思路：
考虑总方案数为(3^n)，我们考虑怎么求出不合法的方案数。令(dp[i][j])表示到第(i)个石头，两条短边和为(j)的方案数
但是我们注意到，如果(sum)是偶数的话，那么：

1. (R = B = frac{sum}{2})和(B = R = frac{sum}{2})
2. (R = G = frac{sum}{2})和(G = R = frac{sum}{2})
3. (B = G = frac{sum}{2})和(G = B = frac{sum}{2})

贡献会重复算一遍，再(dp)一次，删掉一份贡献即可。

代码：

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 310
const ll p = 998244353; 
int n, a[N];
ll f[N * N], g[N * N], all;

int main() {
	while (scanf("%d", &n) != EOF) {
		ll sum = 0, mid;
		all = 1;
		for (int i = 1; i <= n; ++i) {
			scanf("%d", a + i);
			sum += a[i];	
			all = (all * 3) % p; 
		}
		mid = sum / 2;  
		memset(f, 0, sizeof f);  
		f[0] = 1;   
		for (int i = 1; i <= n; ++i) {
			for (int j = sum - a[i]; j >= 0; --j) {
				f[j + a[i]] = (f[j + a[i]] + f[j] * 2 % p) % p;  
			}		 
		}
		
		ll res = 0; 
		for (int i = 0; i <= mid; ++i) {      
			res = (res + f[i]) % p;
		}
		if (sum % 2 == 0) {
			memset(g, 0, sizeof g);
			g[0] = 1;
			for (int i = 1; i <= n; ++i) {
				for (int j = sum - a[i]; j >= 0; --j) {
					g[j + a[i]] = (g[j + a[i]] + g[j]) % p;
				}
			}
			res = (res - g[mid] + p) % p;    
		}
		printf("%lld
", (all - (res * 3) % p + p) % p); 
	}
	return 0;
}