CodeForces 402D Upgrading Array (数学+DP)

题意：给出一个数列，可以进行一种操作将某一个前缀除去他们的gcd,有一个函数f(x)，f(1) = 0 , f(x) = f(x/p)+1,f(x) = f(x/p)-1(p是坏素数)，

求 sum(f[a[i]]) 的最大值。

析：因为f(1) = 0,否则如果是好素数，那么就加一，如果是坏素数就减一，很明显每个数 f(a[i]) 的值就是好素数的数目，送去坏素数的数目，

然后求总的和，这样可以预处理出所有的 gcd，好素数的个数，坏素数的个数，然后dp[i] 表示 sum(f(a[i])) 前 i 个的和最大值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5000 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> prime;
bool vis[(int)1e5+5];
int a[maxn];
void init(){
  int t = (int)sqrt(1e9 + 0.5);
  for(int i = 2; i <= t; ++i)  if(!vis[i]){
    prime.push_back(i);
    for(int j = i*i; j <= t; j += i)  vis[j] = true;
  }
}

int dp[maxn];
int good[maxn];
int bad[maxn];
int gg[maxn];
int gb[maxn];
int g[maxn];
map<int, bool> mp;

void solve(int i, int t, int *good, int *bad){
  for(int j = 0; j < prime.size() && t > 1; ++j) if(t % prime[j] == 0){
    if(mp[prime[j]]){
      while(t % prime[j] == 0){
        t /= prime[j];
        ++bad[i];
      }
    }
    else while(t % prime[j] == 0){
      t /= prime[j];
      ++good[i];
    }
  }
  if(t > 1 && mp[t])  ++bad[i];
  else if(t > 1)  ++good[i];
}

int main(){
  init();
  scanf("%d %d", &n, &m);
  for(int i = 1; i <= n; ++i){
    scanf("%d", a+i);
    g[i] = gcd(g[i-1], a[i]);
  }
  for(int i = 0; i < m; ++i){
    int x;
    scanf("%d", &x);
    mp[x] = true;
  }

  for(int i = 1; i <= n; ++i){
    solve(i, a[i], good, bad);
    good[i] += good[i-1];
    bad[i] += bad[i-1];
    solve(i, g[i], gg, gb);
  }

  for(int i = 0; i <= n; ++i)  dp[i] = -INF;
  dp[0] = 0;
  for(int i = 1; i <= n; ++i)
    for(int j = 0; j < i; ++j)
      dp[i] = max(dp[i], dp[j]+good[i]-good[j]+bad[j]-bad[i]+max(0, (i-j)*(gb[i]-gg[i])));

  printf("%d\n", dp[n]);
  return 0;
}