包子凑数

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题意

给 $n(1\le n\le 100)$ 个数$(1\le a_i\le 100)$，问由它们不能组合(加法)成的数(正整数)的个数，若有无穷个输出”INF”。

题解

若它们的公共 gcd 不为1，输出 INF，构造一个小顶堆，每次新加的值为堆顶的值加$a_i$ ，弹出堆顶，可证最大不可表示的值不会超过1e4，所以当堆顶大于1e4的时候跳出循环。

代码

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#include <bits/stdc++.h>
using namespace std;
#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 fro0(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 oper(type) bool operator <(const type y)const
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
void in(int &a,int &b,int &c,int &d){scanf("%d%d%d%d",&a,&b,&c,&d);}
void in(int &a,int &b,int &c){scanf("%d%d%d",&a,&b,&c);}
void in(int &a,int &b){scanf("%d%d",&a,&b);}
void in(ll &a){scanf("%lld",&a);}
void in(int &a){scanf("%d",&a);}
void in(char *s){scanf("%s",s);}
void in(db &a){scanf("%Lf",&a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 1.1e5;
ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps; }
int db_cmp(db a, db b){ return sign(a-b); }

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}

priority_queue<int,vii,greater<int> >que;
bool vis[11000];
int main() {
#ifdef PerpEternal
    // 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);
    int n,a[110];
    mem0(vis);
    in(n);
    fro0(i,n)in(a[i]);
    if(n==1){
        if(a[0]==1)puts("0");
        else puts("INF");
    }else{
        int gc=a[0];
        for1(i,n-1)gc=gcd(gc,a[i]);
        if(gc==1){
            int ans=0;
            sort(a,a+n);
            fro0(i,n){
                vis[a[i]]=1;
                que.push(a[i]);
            }
            while(1){
                int to=que.top();
                que.pop();
                if(to>=1e4)break;
                fro0(i,n){
                    int now=to+a[i];
                    if(now<=1e4&&vis[now]==0){
                        vis[now]=1;
                        que.push(now);
                    }
                }
            }
            for1(i,1e4)if(!vis[i])ans++;
            printf("%d\n",ans);
        }else puts("INF");
    }
    return 0;
}