A Question of Ingestion(Dp)
阿新 • • 發佈:2018-09-03
ger cti some ngs arch can rom def ase
提交: 95 解決: 26
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Stan is waiting to hear what will be served each hour of the barbecue as he realizes that the menu will determine when and how often he should refrain from eating. For example, if the barbecue lasts 5 hours and the courses served each hour have calories 800, 700, 400, 300, 200 then the best strategy when m = 900 is to eat every hour for a total consumption of 800 + 600 + 400 + 266 + 177 = 2 243 calories. If however, the third course is reduced from 400 calories to 40 calories (some low-calorie celery dish), then the best strategy is to not eat during the third hour — this results in a total consumption of 1 900 calories. The prospect of all this upcoming food has got Stan so frazzled he can’t think straight. Given the number of courses and the number of calories for each course, can you determine the maximum amount of calories Stan can eat?
A Question of Ingestion
時間限制: 1 Sec 內存限制: 128 MB提交: 95 解決: 26
[提交] [狀態] [討論版] [命題人:admin]
題目描述
Stan Ford is a typical college graduate student, meaning that one of the most important things on his mind is where his next meal will be. Fortune has smiled on him as he’s been invited to a multi-course barbecue put on by some of the corporate sponsors of his research team, where each course lasts exactly one hour.Stan is a bit of an analytical type and has determined that his eating pattern over a set of consecutive hours is always very consistent. In the ?rst hour, he can eat up to m calories (where m depends on factors such as stress, bio-rhythms, position of the planets, etc.), but that amount goes down by a factor of two-thirds each consecutive hour afterwards (always truncating in cases of fractions of a calorie). However, if he stops eating for one hour, the next hour he can eat at the same rate as he did before he stopped. So, for example, if m = 900 and he ate for ?ve consecutive hours, the most he could eat each of those hours would be 900, 600, 400, 266 and 177 calories, respectively. If, however, he didn’t eat in the third hour, he could then eat 900, 600, 0, 600 and 400 calories in each of those hours. Furthermore, if Stan can refrain from eating for two hours, then the hour after that he’s capable of eating m calories again. In the example above, if Stan didn’t eat during the third and fourth hours, then he could consume 900, 600, 0, 0 and 900 calories.Stan is waiting to hear what will be served each hour of the barbecue as he realizes that the menu will determine when and how often he should refrain from eating. For example, if the barbecue lasts 5 hours and the courses served each hour have calories 800, 700, 400, 300, 200 then the best strategy when m = 900 is to eat every hour for a total consumption of 800 + 600 + 400 + 266 + 177 = 2 243 calories. If however, the third course is reduced from 400 calories to 40 calories (some low-calorie celery dish), then the best strategy is to not eat during the third hour — this results in a total consumption of 1 900 calories. The prospect of all this upcoming food has got Stan so frazzled he can’t think straight. Given the number of courses and the number of calories for each course, can you determine the maximum amount of calories Stan can eat?
輸入
Input starts with a line containing two positive integers n m (n ≤ 100, m ≤ 20 000) indicating the number of courses and the number of calories Stan can eat in the ?rst hour, respectively. The next line contains n positive integers indicating the number of calories for each course.
輸出
Display the maximum number of calories Stan can consume.
樣例輸入
5 900
800 700 400 300 200
樣例輸出
2243
來源/分類
ecna2017
思路:
100的範圍, 顯然可以 n方 dp。
設計dp狀態,dp(i,j)表示到第 i 天,連續吃了 j 天。然後狀態瞎xx轉移。
代碼如下:
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=110; ll dp[maxn][maxn],a[maxn],f[maxn]; ll n,m; int main(){ scanf("%lld %lld",&n,&m); f[View Code0]=m; for (int i=1; i<=n; i++) { scanf("%lld",&a[i]); f[i]=f[i-1]*2/3; dp[i][0]=min(a[i],m); } for (int i=1; i<=n; i++){ for (int j=0; j<=n; j++){ for (int k=1; k<=n && i+k<=n&&k<=2; k++) dp[i+k][j+2-k]=max(dp[i+k][j+2-k],dp[i][j]+min(f[j+2-k],a[i+k])); dp[i+3][0]=max(dp[i+3][0],dp[i][j]+min(a[i+3],m)); } } ll ans=-1; for (int i=0; i<=n; i++) ans=max(ans,dp[n][i]); printf("%lld\n",ans); return 0; }
A Question of Ingestion(Dp)