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题意：给a个1、b个2、c个5，求不能构成最小的数
思路： 先求1能构成的所有数，2能构成的所有数，5能构成的所有数，它们的方法数显然都是1，现在考虑把3者结合在一起，由于结果为和的形式，而又是循环加的，所以考虑用多项式来表示状态，然后进行两次卷积运算就行了。
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 73 74 75 76 77 | #include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <vector>#include <ctime>#include <deque>#include <queue>#include <algorithm>#include <map>#include <cmath>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 pair<int, int> pii;typedef long long ll;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[12345], b[12345];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);#endif // ONLINE_JUDGE int n1, n2, n3; while (cin >> n1 >> n2 >> n3, n1 + n2 + n3 > 0) { fillchar(a, 0); fillchar(b, 0); for (int i = 0; i <= n1; i ++) { for (int j = 0; j <= 2 * n2; j += 2) { a[i + j] ++; } } int sz = n1 + 2 * n2; for (int i = 0; i <= sz; i ++) { for (int j = 0; j <= 5 * n3; j += 5) { b[i + j] += a[i]; } } for (int i = 0; ; i ++) { if (!b[i]) { cout << i << endl; break; } } } return 0;} |
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原文地址：http://www.cnblogs.com/jklongint/p/4701327.html
