【DP】HDOJ 5291 Candy Distribution

n^4의 DP...그리고 2차 접두사와 1차원 최적화...그리고 넘길 수 있어...

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 200005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL; 
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

LL dp[2][maxn];
int a[maxn];
LL sum[maxn];
int n;

void odd(int st, int t, int delta)
{
	sum[st] += delta;
	sum[st + t] -= 2 * delta;
	sum[st + t + t] += delta;
}

void even(int st, int t, int delta)
{
	sum[st] += delta;
	sum[st + t] -= delta;
	sum[st + t + 2] -= delta;
	sum[st + (t+1) * 2] += delta;
}

void work()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
	
	int base = 50000, now = 0, pre = 1;
	int limit = 0;
	memset(dp, 0, sizeof dp);
	dp[now][base] = 1;
	for(int k = 1; k <= n; k++) {
		now ^= 1, pre ^= 1;
		int val = a[k];
		limit += val;
		for(int i = base - limit - 5; i <= base + limit + 5; i++) dp[now][i] = sum[i] = 0;
		for(int i = base - limit - 5; i <= base + limit + 5; i++) {
			if(dp[pre][i] == 0) continue;
			int len = val / 2 + 1;
			odd(i - 2 * len, 2 * len, dp[pre][i]);
			len = (val - 1) / 2 + 1;
			even(i - 2 * len - 1, 2 * len, dp[pre][i]);
		}
		
		LL t1 = 0, t2 = 0;
		for(int i = base - limit - 4; i <= base + limit + 4; i += 2) {
			t2 = (t2 + sum[i]) % mod;
			dp[now][i] = (dp[now][i] + t1) % mod;
			t1 = (t1 + t2) % mod;
		}
		t1 = t2 = 0;
		for(int i = base - limit - 3; i <= base + limit + 3; i += 2) {
			t2 = (t2 + sum[i]) % mod;
			dp[now][i] = (dp[now][i] + t1) % mod;
			t1 = (t1 + t2) % mod;
		}
	}
	
	printf("%lld
", (dp[now][base] % mod + mod) % mod);
}

int main()
{
	int _;
	scanf("%d", &_);
	while(_--) work();
	
	return 0;
}