HDU-4651

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题意

整数拆分，$(1\le n\le 10^5)$

题解

使用五边形定理

其中 ，注意两个条件要分开判断，有大于0的就加上相应的f，不是两个同时成立或者不成立

代码

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#include <bits/stdc++.h>
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define inlld(lld) scanf("%lld",&lld)
typedef unsigned long long ull;
#define inlf(f) scanf("%lf",&f)
#define ind(d) scanf("%d",&d)
#define ins(s) scanf("%s",s)
#define inf 0x3f3f3f3f
typedef long long ll;
#define pi acos(-1.0)
#define mod (int)(1e9+7)
#define N (int)(1.1e5)
using namespace std;
int p[N];
int main() {
#ifndef ONLINE_JUDGE
  // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
  // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat","w",stdout);
#endif
  //ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
  p[0]=1;
  for1(i,100005){
    ll fl=1,an=0,flag=1;
    while(1){
      ll tmp=fl*(3*fl-1)/2;
      if(i-tmp<0)break;
      an=(an+flag*p[i-tmp]+mod)%mod;
      tmp=fl*(3*fl+1)/2;
      if(i-tmp<0)break;
      an=(an+flag*p[i-tmp]+mod)%mod;
      fl++;
      flag*=-1;
    }
    p[i]=an;
   // printf("%d\n",tmp);
  }
  int t,n;
  ind(t);
  while(t--){
    ind(n);
    printf("%d\n",p[n]);
  }
  return 0;
}