CosmoCraft
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 4257
64-bit integer IO format: %I64d Java class name: Main
In the two-player game CosmoCraft you manage an economy in the hopes of producing an army capable of defeating your opponent. You manage the construction of workers, production facilities, and army units; the game revolves around balancing the resources you allocate to each. The game progresses in turns.
1. Workers give you income at the rate of 1 dollar per turn.
2. Production facilities let you produce either an army unit or a worker for the cost of 1 dollar. (only 1 army unit or worker can be produced per turn per facility)
3. It costs 1 dollar to create a production facility.
4. Your army, of course, lets you fight against your opponent.
You start off with n workers and k production facilities. The game progresses in turns – at each turn, you can spend the income you get from your workers on a mixture of workers, army, and creating production facilities. Workers produced this round do not give you income until the next round; likewise, production facilities do not become active until the next round. Any unspent income from the current round carries over to the next.
At the end of a round, you can take the total army you’ve produced and attack your opponent; if you have strictly more units than your opponent, the opponent loses immediately, and you retain the difference of the army sizes. Otherwise, your army is crushed and your opponent is left with the difference of the army sizes. (it would be wise for him to counter-attack after this, but you don’t lose immediately at least). The game ends after t turns, at which point both players will usually attack with the larger army reigning victorious.
You’re playing against your friend, and since you’ve played against him so many times you know exactly what he’s going to spend his money on at every turn, and exactly when he’s going to attack. Knowing this, you’ve decided that the best strategy is to play defensively – you just want to survive every attack, and amass as large an army in the meantime so you can counterattack (and hopefully win) at the end of the game.
What’s the largest army you can have at the end of the game, given that you must survive all your friend’s attacks?
Input
There will be several test cases in the input. Each test case will begin with a line with three integers:
n k t
where n (1≤n≤100) is the number of workers you start with, k (1≤k≤100) is the number of production facilities you have at the start, and t(1≤t≤10,000) is the number of turns. On the next line will be t-1 integers, ai (0≤ai≤Max signed 64-bit integer), separated by single spaces. The ith integer indicates the strength of the attack (that is, the number of army units your opponent is using in that attack) on turn i. The input will end with a line with three 0s.
Hint
Huge input, please ues c++.
Output
For each test case output a single integer indicating the maximum number of armies you could have at the end of the game. Output -1 if it is impossible to survive. Output each integer on its own line, with no spaces, and do not print any blank lines between answers. While it is possible for some inputs to generate unreasonably large answers, all judge inputs yield answers which will fit in a signed 64-bit integer.
Sample Input
8 4 6 22 6 10 14 0 4 3 3 0 0 6 9 7 0 0 11 0 7 0 0 0 0
Sample Output
-1 11 101
Source
The University of Chicago Invitational Programming Contest 2012
解题：转自他人
题目大意
介绍一款名叫CosmoCraft的双人回合制策略游戏（但网上找不到这种游戏），每一回合玩家可以：１。花费1元/个购买工人，2。花费1元/个购买士兵，3。花费1元/个购买兵营。每回合1个工人可以挣1元，但钱会在下一回合给你，每个兵营每一回合只能造一个工人或士兵，同样的每一回合建造的兵营将在下一回合起可用，每一回合造的工人或士兵当前回合起即可使用。游戏初始给你n个工人和k个兵营以及n元，游戏共t回合，前t-1回合每一回合将会有a[i]个敌方士兵回来攻击你，你必须能抵挡住每一回合的进攻，即在每一回合拥有比进攻你的敌方军队更多的士兵，每次战争士兵数多的一方取胜，并剩下双方交战士兵的差值的士兵数，而失败一方则一兵不剩，请你使用最优策略使得你在第t回合的反击中拥有最多的士兵，并输出最大值。
贪心策略：
１。每一回合必须花完所有的钱。
２。一个士兵不可能连续存在两回合以上（超过两回合说明他没有打仗，可以花１元在第一回合造工人，在第二回合用工人赚的钱造兵营，第三回合再用工人赚的钱和造的兵营造士兵）。
３。每回合在保证生存的前提下，先造尽量多的工人，能造多少造多少，剩下的钱再造兵营。
４。根据策略二，第i(1<=i<=t)回合的士兵只能由第i回合或第i-1回合造，若第i回合可以造出d[i]个士兵，就在第i回合造，否则需要第i-1回合的支援，第i-1回合最多支援第i回合k[i-1]-u[i-1]个士兵（在w[i-1]>k[i-1]的情况下）否则不能抵御第i回合的进攻。（k[i]代表第i回合的兵营数，n[i]代表第i回合的工人数，u[i]代表第i回合所需士兵数）。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 100100;
18 LL d[maxn],n,k,t,u,theMin;
19 int main() {
20 int i,j;
21 bool flag;
22 while(scanf("%I64d %I64d %I64d",&n,&k,&t),n||k||t){
23 for(i = 1; i < t; i++) scanf("%I64d",d+i);
24 d[t] = 0;
25 if(min(n,k) < d[1]){puts("-1");continue;}
26 u = d[1];
27 flag = false;
28 for(i = 1; i < t; i++){
29 if(n + min(n,k) - u < d[i+1]) {flag = true;break;}//利用上次剩余的
30 //造一些 还不够去死
31 theMin = min(n,k);//利用n个人和k个军营最多造的人数
32 k += n - theMin;//当n > k时，剩余的钱造军营
33 n += theMin - u;//造了theMin个人，然后去打仗，死了一些，最后剩余
34 if(d[i+1] > k){//下一轮进攻军营不够用
35 n -= d[i+1]-k;//用一些工人去换d[i+1]-k个人，
36 u = k;//已经死了这么多个，还要死k个才行
37 }else u = d[i+1];
38 }
39 if(flag) {puts("-1");continue;}
40 if(t == 1) printf("%I64d\n",u < k ? u:k);
41 else printf("%I64d\n",n);
42 }
43 return 0;
44 }
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