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; }