题目链接：点击打开链接
题意：
给定n种面额，每次ATM机能吐k张纸钞。（且每次ATM机只能吐至多2种面额)
q个询问，，每次回答若一次能吐出该面额的最少纸张数量，若不能则输出-1
O(n*k*k*logn)能过
#include<iostream>#include<stdio.h>#include<string.h>#include <algorithm>#include<string>#include<set>#include<vector>#include<queue>#include<math.h>template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c – '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c – '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if (x>9) pt(x / 10);putchar(x % 10 + '0');}using namespace std;typedef long long ll;const int N = 5005;const int inf = 1000000;int n, k;ll a[N];int hehe; int dd;set<ll>myset;ll work(ll x){ll ans = inf;for (int i = 1; i <= n; i++){if (x%a[i] == 0 && x / a[i] <= k)ans = min(ans, x / a[i]);for (int j = 0; j <= k; j++){ll y = x – a[i] * j;for (int z = 1; z <= k; z++){if (z + j > k)break;if (y%z == 0 && myset.count(y / z))ans = min(ans, (ll)z + j);}}}if (ans == inf)return -1;return ans;}int main(){while (cin >> n >> k){myset.clear();for (int i = 1; i <= n; i++)rd(a[i]), myset.insert(a[i]);int q; rd(q);while (q–){ll x; rd(x);pt(work(x)); puts("");}}return 0;}
爱上一个人的时候，总会有点害怕，怕得到他;怕失掉他。
