Codeforces 515C - Drazil and Factorial (贪心)

题意

给出一些规则,输出最大的符合规则的数。

思路

数越大,说明数的个数越多。
所以能分解就分解

把4、6、8、9分解成质因数相乘,然后从7开始。

代码

#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
const int hash_size = 4e5 + 10;
int cases = 0;
typedef pair<int, int> pii;
 
vector<int> ans;
int cnt[10];
 
void Resolve(int k)
{
    for (int i = 2; i <= k; i++) cnt[i]++;
    if (cnt[9]) cnt[3] += 2, cnt[9] = 0;
    if (cnt[6]) cnt[2]++, cnt[3]++, cnt[6] = 0;
    if (cnt[4]) cnt[2] += 2, cnt[4] = 0;
    if (cnt[8]) cnt[2] += 3, cnt[8] = 0;;
}
 
int main()
{
    //ROP;
    int len;
    scanf("%d%*c", &len);
    for (int i = 0; i < len; i++)
    {
        char tmp;
        tmp = getchar();
        Resolve(tmp-'0');
    }
    for (int i = 7; i >= 2; i--)
    {
        if (!cnt[i]) continue;
        int k = cnt[i];
        for (int j = i; j >= 2; j--)
        {
            if (cnt[j] == 0)
            {
                if (j == 4) cnt[2] -= 2*k;
                if (j == 6) cnt[3] -= k, cnt[2] -= k;
            }
            else cnt[j] -= k;
        }
        for (int j = 0; j < k; j++) ans.PB(i);
    }
    for (int i = 0; i < SZ(ans); i++) printf("%d", ans[i]);
    return 0;
}

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