poj1952求最长递减子序列和最长子序列的种数
2014-03-08 16:20
337 查看
求最长递减子序列和最长子序列的种数。难在求种数
不需要使用高精度或者int64
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:
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:
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:
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
Sample Output
不需要使用高精度或者int64
BUY LOW, BUY LOWER
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 5490 | Accepted: 1848 |
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
#include <iostream> #define N 5005 using namespace std; int a ; int dp ; int times ; int main() { int n; while(cin>>n) { for(int i=1;i<=n;i++) { cin>>a[i]; dp[i]=times[i]=1; } for(int i=1;i<=n;i++)//求以a[i]为终止时子序列的长度 { for(int j=i-1;j>0;j--) { if(a[i]<a[j]) { if(dp[i]<dp[j]+1) { dp[i]=dp[j]+1; times[i]=times[j]; } else if(dp[i]==dp[j]+1) { times[i]+=times[j]; } } if(a[i]==a[j]) { if(dp[i]==1)//防止重复 //dp[i]==1,a[i],a[j]前面接的一样,只能算作一次 times[i]=0; break; } } } int len=0,t=0; for(int i=1;i<=n;i++) if(dp[i]>len) len=dp[i]; for(int i=1;i<=n;i++) if(dp[i]==len)//统计次数 t+=times[i]; cout<<len<<" "<<t<<endl; } }
相关文章推荐
- Native wifi API使用
- "码代码"微信号今日上线,为互联网同仁提供最前沿咨询
- Native wifi API使用
- Linux入门之pxe网络自动安装系统----“高级的自动化”
- [Qt] 循序容器(QVector、QLinkedList、QList...)
- 自定义一个ImageSwitcher
- C#中的反射 Assembly.Load() Assembly.LoadFrom()
- 开发资源收藏
- 设计模式六大原则
- 2007偶数的平方和和奇数的立方和
- 一个很纠结的问题
- visual studio 生成后事件 Post-Build Event
- 开发资源收藏
- git+github+markdown+jekyll搭建个人博客
- 科大讯飞语音合成Demo讲讲(做做玩,超简)
- json格式序列化注意点
- Android关于OnTouch 和OnClick同时调用冲突的解决方案
- IDC:今年全球平板销量2.5亿部 涨幅显著放缓
- 网络编程之 Socket函数 (一)
- VPC/VM/VBOX安装GHOST版的无法启动系统