HDU 4651 Partition(五边形数)

题意:

$N\le 10^5,求整数N的划分数是多少,答案对10^9+7取模$

分析:

$这个东西就是五边形数,当然n^{1.5}的dp也可以,太神了不懂$
$所以直接公式搞就好辣,时间复杂度也是O(n^{1.5})$
$f(0)=1,f(n)=\displaystyle\sum_{k=1}(-1)^{k+1}\cdot \left( f(n-{k(3k-1)\over 2}) + f(n-{k(3k+1)\over 2})\right)$

代码:

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//
// Created by TaoSama on 2016-03-11
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
//五边形数: f(n) = sum( (-1)^(k+1) [ P(n-k(3k-1)/2) + P(n-k(3k+1)/2)] )
int n, f[N];
void add(int& x, int y) {
x += y;
x %= MOD;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
f[0] = 1;
for(int i = 1; i <= 1e5; ++i) {
int sum = 0, delta = 1;
for(int j = 1; ; ++j, delta = -delta) {
int x = i - j * (3 * j - 1) / 2;
int y = i - j * (3 * j + 1) / 2;
if(x < 0 && y < 0) break;
if(x >= 0) add(sum, delta * f[x]);
if(y >= 0) add(sum, delta * f[y]);
}
f[i] = (sum % MOD + MOD) % MOD;
}
int t; scanf("%d", &t);
while(t--) {
int n; scanf("%d", &n);
printf("%d\n", f[n]);
}
return 0;
}


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