Codeforces 510D - Fox And Jumping (DP)
题意
给出几个数字和他们的权值,要求选择k个数,他们能凑出所有的自然数,问最小权。
思路
最终目标是要凑出1. 几个数能凑的最小的正整数就是他们的GCD。递推一下。
代码
#include <stack>
#include <cstdio>
#include <list>
#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 = 300 + 10;
const int MOD = 29;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
pii arr[MAXN];
map<int, int> mp;
int main()
{
//ROP;
int n;
scanf("%d", &n);
mp[0] = 0;
for (int i = 0; i < n; i++) scanf("%d", &arr[i].X);
for (int i = 0; i < n; i++) scanf("%d", &arr[i].Y);
for (int i = 0; i < n; i++)
{
map<int, int>::iterator it = mp.begin();
map<int, int> tmp;
for (; it != mp.end(); it++)
{
int num = __gcd(it->X, arr[i].X);
int cost = it->Y + arr[i].Y;
if (!mp.count(num) || mp[num] > cost)
{
if (tmp.count(num) && tmp[num] < cost) continue;
tmp[num] = cost;
}
}
for (it = tmp.begin(); it != tmp.end(); it++) mp[it->X] = it->Y;
}
printf("%d\n", mp.count(1) ? mp[1] : -1);
return 0;
}