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

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;
}