POJ 1952
BUY LOW, BUY LOWER
Description
The advice to "buy low" is half the formula to success in the bovine stock market.To be considered a great investor you must also follow this problems' advice:
"Buy low; buy lower"
Each time you buy a stock, you must purchase it at a lower price than the previous time you bought it. The more times you buy at a lower price than before, the better! Your goal is to see how many times you can continue purchasing at ever lower prices.
You will be given the daily selling prices of a stock (positive 16-bit integers) over a period of time. You can choose to buy stock on any of the days. Each time you choose to buy, the price must be strictly lower than the previous time you bought stock. Write a program which identifies which days you should buy stock in order to maximize the number of times you buy.
Here is a list of stock prices:
Day 1 2 3 4 5 6 7 8 9 10 11 12
Price 68 69 54 64 68 64 70 67 78 62 98 87
The best investor (by this problem, anyway) can buy at most four times if each purchase is lower then the previous purchase. One four day sequence (there might be others) of acceptable buys is:
Day 2 5 6 10
Price 69 68 64 62
Input
* Line 1: N (1 <= N <= 5000), the number of days for which stock prices are given
* Lines 2..etc: A series of N space-separated integers, ten per line except the final line which might have fewer integers.
Output
Two integers on a single line:
* The length of the longest sequence of decreasing prices
* The number of sequences that have this length (guaranteed to fit in 31 bits)
In counting the number of solutions, two potential solutions are considered the same (and would only count as one solution) if they repeat the same string of decreasing prices, that is, if they "look the same" when the successive prices are compared. Thus, two different sequence of "buy" days could produce the same string of decreasing prices and be counted as only a single solution.
Sample Input
12
68 69 54 64 68 64 70 67 78 62
98 87
Sample Output
4 2
/翻译
“买低”的建议是牛股市成功的一半。要想成为一个伟大的投资者，你也必须遵循这个问题的建议：
“低买低买”
每次你买一只股票时，你必须以较低的价格购买它。你以较低的价格购买的次数越多，越好！你的目标是看看你能以更低的价格继续购买多少次。
在一段时间内，你将得到股票（正数16位整数）的每日销售价格。你可以选择在任何一天购买股票。每一次你选择购买，价格必须严格低于你买股票的前一次。写一个程序，确定哪些天你应该买股票，以最大限度地增加你买的次数。
这是一份股票价格表：
第1天2，3，4，5，6，7，8，9，10，11，12
价格68，69，54，64，68，64，70，67，78，62，98，87
最好的投资者（不管怎么说），如果每次购买都比上次购买的低，那么最多可以购买四次。一个四天的序列（可能还有其他）可接受的购买是：
第2天5 6 10
价格69 68 64 62
思路
最长下降子序列
加上统计可能的情况
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cmath> 6 #include<algorithm> 7 8 using namespace std; 9 10 int n,mxl,ans,a[6000],f[6000],g[6000]; 11 12 int read() 13 { 14 int x=0,f=1; 15 char ch=getchar(); 16 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 17 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 18 return x*f; 19 } 20 int main() 21 { 22 n=read(); 23 for(int i=1;i<=n;i++)a[i]=read(); 24 for(int i=1;i<=n;i++) 25 { 26 bool flg=false; 27 for(int j=i-1;j>=1;j--) 28 { 29 if(a[j]==a[i]){flg=true;break;} 30 if(a[j]>a[i]) 31 { 32 if(f[j]+1>f[i]) 33 { 34 f[i]=f[j]+1; 35 g[i]=g[j]; 36 } 37 else if(f[j]+1==f[i])g[i]+=g[j]; 38 } 39 } 40 if(!flg) 41 { 42 f[i]=max(f[i],1); 43 g[i]=max(g[i],1); 44 } 45 mxl=max(mxl,f[i]); 46 } 47 for(int i=1;i<=n;i++) 48 { 49 if(f[i]==mxl)ans+=g[i]; 50 } 51 printf("%d %d",mxl,ans); 52 system("pause"); 53 return 0; 54 }