Codeforces Round #556 (Div. 2) - C. Prefix Sum Primes（思维）

Problem  Codeforces Round #556 (Div. 2) - D. Three Religions

Time Limit: 1000 mSec

Problem Description

Input

Output

Sample Input

5
1 2 1 2 1

Sample Output

1 1 1 2 2

题解：这个题有做慢了，这种题做慢了和没做出来区别不大。。。

　　读题的时候脑子里还意识到素数除了2都是奇数，读完之后就脑子里就只剩欧拉筛了，贪心地构造使得前缀和是连续的素数，那实现就很简单了，将素数序列的差分序列求出来，不断凑出差分序列的每个数即可，但是之后想想，除了2 和 3，每个的间隔不都是偶数么，肯定是连续的2呀，费劲算差分序列干什么，直接先放2再放1不就行了（特殊处理一下2 和 3 即可），写着写着还误以为要输出下标，临时改了改，等到测样例的时候发现是输出1、2，暴风哭泣。

  1 #include <bits/stdc++.h>
  2
  3 using namespace std;
  4
  5 #define REP(i, n) for (int i = 1; i <= (n); i++)
  6 #define sqr(x) ((x) * (x))
  7
  8 const int maxn = 400000 + 10000;
  9 const int maxm = 200000 + 100;
 10 const int maxs = 10000 + 10;
 11
 12 typedef long long LL;
 13 typedef pair<int, int> pii;
 14 typedef pair<double, double> pdd;
 15
 16 const LL unit = 1LL;
 17 const int INF = 0x3f3f3f3f;
 18 const double eps = 1e-14;
 19 const double inf = 1e15;
 20 const double pi = acos(-1.0);
 21 const int SIZE = 100 + 5;
 22 const LL MOD = 1000000007;
 23
 24 LL n;
 25 LL a[maxn];
 26 int cnt[5];
 27 LL cnt1, cnt2;
 28 LL tot, prime[maxn];
 29 bool is_prime[maxn];
 30
 31 void Euler()
 32 {
 33     memset(is_prime, true, sizeof(is_prime));
 34     is_prime[0] = is_prime[1] = false;
 35     for (LL i = 2; i < maxn; i++)
 36     {
 37         if (is_prime[i])
 38         {
 39             prime[tot++] = i;
 40         }
 41         for (LL j = 0; j < tot && i * prime[j] < maxn; j++)
 42         {
 43             is_prime[prime[j] * i] = false;
 44             if (i % prime[j] == 0)
 45             {
 46                 break;
 47             }
 48         }
 49     }
 50 }
 51
 52 vector<int> ans;
 53 queue<int> que[3];
 54
 55 int main()
 56 {
 57     ios::sync_with_stdio(false);
 58     cin.tie(0);
 59     //freopen("input.txt", "r", stdin);
 60     //freopen("output.txt", "w", stdout);
 61     Euler();
 62     cin >> n;
 63     int x;
 64     LL sum = 0;
 65     for (int i = 1; i <= n; i++)
 66     {
 67         cin >> x;
 68         que[x].push(i);
 69         cnt[x]++;
 70         sum += x;
 71     }
 72     cnt1 = cnt[1], cnt2 = cnt[2];
 73     LL pre = 0;
 74     for (int i = 0; i < tot && prime[i] <= sum; i++)
 75     {
 76         LL tmp = prime[i] - pre;
 77         LL x = tmp / 2;
 78         x = min(x, cnt2);
 79         if (tmp - x * 2 <= cnt1)
 80         {
 81             pre = prime[i];
 82             for (int j = 0; j < x; j++)
 83             {
 84                 ans.push_back(2);
 85             }
 86             for (int j = 0; j < tmp - x * 2; j++)
 87             {
 88                 ans.push_back(1);
 89             }
 90             cnt2 -= x;
 91             cnt1 -= (tmp - x * 2);
 92         }
 93     }
 94     for(int i = 0; i < ans.size(); i++)
 95     {
 96         cout << ans[i] << " ";
 97     }
 98     while(cnt1--)
 99     {
100         cout << 1 << " ";
101     }
102     while(cnt2--)
103     {
104         cout << 2 << " ";
105     }
106     return 0;
107 }

原文地址：https://www.cnblogs.com/npugen/p/10798353.html