CodeForces 524C The Art of Dealing with ATM (二分)

时间：2024-10-22 14:06:14

题意：给定 n 种不同的钞票，然后用q个询问，问你用最多k张，最多两种不同的钞票能不能组成一个值。

析：首先如果要求的值小点，就可以用DP，但是太大了，所以我们考虑一共最多有n * k种钞票，如果每次都挨着遍历，时间肯定受不了，

所以我们可以枚举其中一种，然后再用二分查找快速查找另一种，然后不断更新答案。

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<double, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-5;

const int maxn = 100 + 10;

const int mod = 1e6;

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> v[25];

int main(){

  scanf("%d %d", &n, &m);

  for(int i = 0; i < n; ++i){

    int x;

    scanf("%d", &x);

    for(int j = 1; j <= m; ++j)  v[j].push_back(x * j);

  }

  for(int i = 1; i <= m; ++i) sort(v[i].begin(), v[i].end());

  int q;

  scanf("%d", &q);

  while(q--){

    int x;

    scanf("%d", &x);

    int ans = INF;

    if(x == 0){  printf("0\n");  continue; }

    for(int i = 1; i <= m; ++i)

      for(int j = 0; j < v[i].size(); ++j){

        int y = x - v[i][j];

        if(y == 0){ ans = min(ans, i);  break; }

        if(y < 0)  break;

        for(int k = 1; k <= m - i; ++k){

          int pos = lower_bound(v[k].begin(), v[k].end(), y) - v[k].begin();

          if(pos < v[k].size() && v[k][pos] == y){

            ans = min(ans, i+k);

          }

        }

      }

      printf("%d\n", ans == INF ? -1 : ans);

  }

  return 0;

}

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