題意:給n張票的價格pi,每a張票價格的x%用作環保項目1,每b張票價格的y%用作環保項目2,爲了使投入環保項目的總額至少爲k,求出重新排列票的價格後需要賣出的最少的票數
思路:二分票數,用大頂堆來存儲票的價格,顯然a和b的公倍數位置上應該放最大的價格,其次比較x和y,較大的那個對應的位置上放較大價格,最後把剩餘的沒放的位置再放剩下的價格,每次從堆中取價格,判斷當前票數是否滿足價格至少爲k,滿足則減小右邊界,存儲答案,否則增加左邊界,最後答案即爲最小票數。
坑點有long long問題,還有double精度問題,之前把x和y轉化爲double類型的百分比,但這樣使得答案和預想的不一樣,實際上題目指定票的價格爲100的倍數,故直接用票的價格除以100乘上百分率。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5+10;
const int inf = 0x3f3f3f3f;
ll q, n, x, y, a, b, k;
priority_queue<ll> que;
ll lcm(ll x1, ll y1)
{
return x1*y1/__gcd(x1, y1);
}
bool check(ll z)
{
ll cur = 0;
priority_queue<ll> que2 = que;
ll c = lcm(a, b);
if (z >= c) {
for (ll i = c; i <= z; i += c) {
ll t = que2.top(); que2.pop();
cur += t*(x+y)/100;
}
}
if (z >= a) {
for (ll i = a; i <= z; i += a) {
if (i%c != 0) {
ll t = que2.top(); que2.pop();
cur += t*x/100;
}
}
}
for (ll i = b; i <= z; i += b) {
if (i%c != 0) {
ll t = que2.top(); que2.pop();
cur += t*y/100;
}
}
return cur >= k;
}
int main()
{
cin >> q;
while (q--) {
cin >> n;
while (!que.empty())
que.pop();
for (int i = 0; i < n; i++) {
ll t;
scanf("%lld", &t);
que.push(t);
}
scanf("%lld%lld", &x, &a);
scanf("%lld%lld", &y, &b);
scanf("%lld", &k);
if (x < y) {
swap(x, y);
swap(a, b);
}
ll l = 0, r = 2*n, ans = -1;
while (l <= r) {
ll mid = (l+r)>>1;
if (check(mid)) {
r = mid - 1;
ans = mid;
}
else
l = mid + 1;
}
if (ans > n)
ans = -1;
cout << ans << "\n";
}
return 0;
}