usaco Beef McNuggets
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1.没有得不到的数,最小的盒子为1
2.得不到的数无穷大,所有数的最大公约数大于1,因为得到的数只能为该公约数的倍数
3.如果存在的话,结果不会超过最大的两个数的最小公倍数
证明见:点击打开链接
4.多重背包
/* ID: daijinq1 PROB:nuggets LANG: C++ */#include <iostream>#include <algorithm>#include <sstream>#include <cstring>#include <cstdio>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)#define FORD(i,a,b) for(i = (a); i > (b); --i)#define FORDE(i,a,b) for(i = (a); i >= (b); --i)#define CLR(a,b) memset(a,b,sizeof(a))const int MAXN = 250000;const int INF = 0x7f7f7f7f;bool vis[200000];int N, T, M;int num[MAXN];int gcd(int a, int b){ return (b == 0) ? a : gcd(b, a % b); }void solve() { int i, j; int a = num[1]; FORE(i, 1, N) { a = gcd(a, num[i]); } if(a != 1 || N == 1) { cout<<"0"<<endl; return ; } vis[0] = true; FORE(j, 1, num[1] * num[2]) FORE(i, 1, N) if(j >= num[i] && vis[j - num[i]]) { vis[j] = true; break; } FORDE(i, num[1] * num[2], 1) { if(!vis[i]) break; } cout<<i<<endl;}void input() { int i; cin>>N; FORE(i, 1, N) cin>>num[i]; sort(num, num + N + 1);}int main() { freopen("nuggets.in", "r", stdin); freopen("nuggets.out", "w", stdout); input(); solve(); return 0;}
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