Codeforces 725E Too Much Money (看题解)

Too Much Money

最关键的一点就是这个贪心可以在sqrt(n)级别算出答案。

因为最多有sqrt(n)个不同的数值加入。

我们可以发现最优肯定加入一个。

然后维护一个当前可以取的最大值， 枚举加入的数来贪心。

#include
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair
#define PLI pair
#define PII pair
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;
const double eps = 1e-6;
const double PI = acos(-1);

int tar, n, tot, a[N], hs[N], cnt[N], mx[N], now[N];
vector<int> vc;

bool check(int add) {
    vc.clear();
    int c = tar;
    int p = mx[c];
    while(c) {
        if(add > c) add = 0;
        while(!now[p] && p) p--;
        p = min(p, mx[c]);
        if(!p && !add) break;
        if(p && hs[p] >= add) {
            vc.push_back(p);
            int v = hs[p], num = min(now[p], c / v);
            now[p] -= num;
            c -= num * v;
        } else {
            c -= add;
            add = 0;
        }
    }
    for(auto& x : vc) now[x] = cnt[x];
    return c;
}

int main() {
    scanf("%d%d", &tar, &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        hs[++tot] = a[i];
    }
    sort(hs + 1, hs + tot + 1);
    tot = unique(hs + 1, hs + tot + 1) - hs - 1;
    for(int i = 1; i <= n; i++) cnt[lower_bound(hs + 1, hs + tot + 1, a[i]) - hs]++;
    for(int i = 1; i <= tot; i++) now[i] = cnt[i];
    for(int i = 1; i <= tar; i++) mx[i] = upper_bound(hs + 1, hs + tot + 1, i) - hs - 1;
    for(int i = 1; i <= tar; i++) {
        if(check(i)) {
            printf("%d\n", i);
            return 0;
        }
    }
    puts("Greed is good");
    return 0;
}

/*
*/

 

转载于:https://www.cnblogs.com/CJLHY/p/10559955.html