题面

Sol

状压一下\(k\)，\(f[S]\)表示用过的硬币集合为\(S\)能买到的物品个数

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(100005);IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int k, coin[_], n, cost[_], sum[_], f[65536], ans = -1, mi[20];int main(RG int argc, RG char* argv[]){    k = Input(); n = Input();    for(RG int i = 1; i <= k; ++i) coin[i] = Input();    for(RG int i = 1; i <= n; ++i){        cost[i] = Input();        sum[i] = sum[i - 1] + cost[i];    }    mi[1] = 1;    for(RG int i = 2; i <= k; ++i) mi[i] = mi[i - 1] << 1;    RG int S = 1 << k;    for(RG int i = 0; i < S; ++i)        for(RG int j = 1; j <= k; ++j){            if(i & mi[j]) continue;            RG int pos = lower_bound(sum + 1, sum + n + 1, coin[j] + sum[f[i]]) - sum;            if(pos > n || sum[pos] > coin[j] + sum[f[i]]) --pos;            f[i | mi[j]] = max(f[i | mi[j]], pos);        }    for(RG int i = 0; i < S; ++i)        if(f[i] == n){            RG int cnt = 0;            for(RG int j = 1; j <= k; ++j)                if(~i & mi[j]) cnt += coin[j];            ans = max(ans, cnt);        }    printf("%d\n", ans);    return 0;}