题意：给一个正整数n，求n的拆分方法数（不考虑顺序）
思路：不妨考虑用1~n来构成n。用多项式表示单个数所有能构成的数，用多项式表示，就相当于卷积运算了。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 | #include <map>#include <set>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <vector>#include <cstdio>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define X first#define Y second#define pb push_back#define mp make_pair#define all(a) (a).begin(), (a).end()#define fillchar(a, x) memset(a, x, sizeof(a))typedef long long ll;typedef pair<int, int> pii;typedef unsigned long long ull;#ifndef ONLINE_JUDGEvoid RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>void print(const T t){cout<<t<<endl;}template<typename F,typename...R>void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}#endiftemplate<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}template<typename T>void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}template<typename T>void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}const double PI = acos(-1.0);const int INF = 1e9 + 7;/* -------------------------------------------------------------------------------- */int a[123], b[123];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);#endif // ONLINE_JUDGE int n; while (cin >> n) { for (int i = 0; i <= n; i ++) a[i] = 1; for (int i = 2; i <= n; i ++) { fillchar(b, 0); for (int j = 0; j <= n; j ++) { for (int k = 0; k <= n / i; k ++) { if (j + k * i > n) break; b[j + k * i] += a[j]; } } memcpy(a, b, sizeof(b)); } cout << a[n] << endl; } return 0;} |