codeforces 883C Downloading B++ 枚举,贪心

Downloading B++

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Only T milliseconds left before the start of well-known online programming contest Codehorses Round 2017.

Polycarp needs to download B++ compiler to take part in the contest. The size of the file is f bytes.

Polycarp's internet tariff allows to download data at the rate of one byte per t0 milliseconds. This tariff is already prepaid, and its use does not incur any expense for Polycarp. In addition, the Internet service provider offers two additional packages:

• download a1 bytes at the rate of one byte per t1 milliseconds, paying p1 burles for the package;
• download a2 bytes at the rate of one byte per t2 milliseconds, paying p2 burles for the package.

Polycarp can buy any package many times. When buying a package, its price (p1 or p2) is prepaid before usage. Once a package is bought it replaces the regular tariff until package data limit is completely used. After a package is consumed Polycarp can immediately buy a new package or switch to the regular tariff without loosing any time. While a package is in use Polycarp can't buy another package or switch back to the regular internet tariff.

Find the minimum amount of money Polycarp has to spend to download an f bytes file no more than in T milliseconds.

Note that because of technical reasons Polycarp can download only integer number of bytes using regular tariff and both packages. I.e. in each of three downloading modes the number of downloaded bytes will be integer. It means that Polycarp can't download a byte partially using the regular tariff or/and both packages.

Input

The first line contains three integer numbers f, T and t0 (1 ≤ f, T, t0 ≤ 107) — size of the file to download (in bytes), maximal time to download the file (in milliseconds) and number of milliseconds to download one byte using the regular internet tariff.

The second line contains a description of the first additional package. The line contains three integer numbers a1, t1 and p1(1 ≤ a1, t1, p1 ≤ 107), where a1 is maximal sizes of downloaded data (in bytes), t1 is time to download one byte (in milliseconds), p1 is price of the package (in burles).

The third line contains a description of the second additional package. The line contains three integer numbers a2, t2 and p2(1 ≤ a2, t2, p2 ≤ 107), where a2 is maximal sizes of downloaded data (in bytes), t2 is time to download one byte (in milliseconds), p2 is price of the package (in burles).

Polycarp can buy any package many times. Once package is bought it replaces the regular tariff until package data limit is completely used. While a package is in use Polycarp can't buy another package or switch back to the regular internet tariff.

Output

Print the minimum amount of money that Polycarp needs to pay to download B++ compiler no more than in T milliseconds. If there is no solution, print the only integer -1.

题解：

枚举时间最短的套餐使用个数，得到剩余F（剩余数据包）以及剩余T（剩余时间）

然后使用最少的第二个套餐的个数，使得下载F量的数据包满足时间不超过剩余T。（相当于一元不等式，直接解就ok了，有一些细节自己考虑一下）

代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL inf = 1e18;LL f,T,t0,a1,t1,p1,a2,t2,p2,resF,resT,ans = inf;int main(){cin>>f>>T>>t0>>a1>>t1>>p1>>a2>>t2>>p2;if(t1>t2) swap(a1,a2),swap(p1,p2),swap(t1,t2);for(int i = 0;(i-1)*a1 <= f;++i){resF = f-min(i*a1,f),resT = T-min(i*a1,f)*t1;if(resF == 0) {if(resT < 0)break;ans = min(ans,i*p1);break;}if(t0*resF>resT && t2*resF>resT) continue;if(t0*resF <= resT) ans = min(ans,i*p1);else{LL x = (resT-t0*resF)/(t2-t0) + ((resT-t0*resF)%(t2-t0)==0?0:1);//x = max(0ll,x);LL pk = (x/a2)+(x%a2==0?0:1);ans = min(ans,pk*p2+i*p1);}}ans = max(ans,(LL)0);printf("%lld\n",ans < inf?ans:-1);return 0;}