51NOD 1259 整数划分 (五边形定理)

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题意

将N分为若干个整数的和，求有多少种不同的划分方式。

思路

几个月前妄图用DP过这题失败了= =，今天队友问起这题才想起来我还没过，看到discuss说用五边形定理赶紧去学习了一波。。。
参考ACdreamer巨巨的博客：http://blog.csdn.net/acdreamers/article/details/12259815

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 7;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;LL ans[maxn];LL temp[maxn];void init(){    int t = 1000;    for (int i = -1000; i <= 1000; i++)        temp[i+t] = i * (3 * i - 1) / 2;    ans[0] = 1;    for (int i = 1; i <= maxn; i++)    {        int sum = 0;        for (int j = 1; j <= i; j++)        {            if (temp[t+j] <= i)            {                if (j & 1) sum = (sum + ans[i-temp[t+j]] + MOD) % MOD;                else sum = (sum - ans[i-temp[t+j]] + MOD) % MOD;            }            else                break;            if (temp[t-j] <= i)            {                if (j & 1) sum = (sum + ans[i-temp[t-j]] + MOD) % MOD;                else sum = (sum - ans[i-temp[t-j]] + MOD) % MOD;            }            else                 break;        }        ans[i] = sum;    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    init();    int n;    while (scanf("%d", &n) != EOF)    {        printf("%lld\n", ans[n]);    }    return 0;}

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