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 87Sample Output
4 2
求最长严格下降子序列的长度以及本质不同的方案数。
先离散化，转换成求最长严格上升的，用L[i]来统计以i结尾的最长长度是多少。
用dp[i]来统计方案数，由于需要的是本质不同的，那么对于两个数a[i]和a[j]来说，
如果a[i]==a[j]，那么我们显然只需要位置靠后的那个，前面那个是不需要统计的，
因为前面能形成的串必然可以被后面那个形成，这样就ok了。
顺便一说，统计方案数不需要大数，因为又要本质不同，又要长度最长，这样的方案数显然不可能很多。
#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<string,string>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 5e3 + 10;
const int read()
{
char ch = getchar();
while (ch<'0' || ch>'9') ch = getchar();
int x = ch - '0';
while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';
return x;
}
int T, n, m, a[N], b[N];
int dp[N], L[N], f[N];
int main()
{
while (scanf("%d", &n) != EOF)
{
rep(i, 1, n) scanf("%d", &a[i]), b[i] = a[i];
sort(b + 1, b + n + 1); m = unique(b + 1, b + n + 1) - b;
int ans = 0, len = 0;
rep(i, 1, n)
{
a[i] = m - (lower_bound(b + 1, b + m, a[i]) - b);
L[i] = 1; dp[i] = 0;
per(j, i - 1, 1) if (a[j] < a[i]) L[i] = max(L[i], L[j] + 1);
len = max(L[i], len);
if (L[i] == 1) { dp[i] = 1; continue; }
rep(j, 1, m) f[j] = 0;
per(j, i - 1, 1)
{
if (a[j] >= a[i] || L[i] != L[j] + 1) continue;
if (f[a[j]]) continue; else f[a[j]] = 1;
dp[i] += dp[j];
}
}
rep(i, 1, m) f[i] = 0;
per(i, n, 1)
{
if (f[a[i]]) continue; else f[a[i]] = 1;
ans += (L[i] == len) * dp[i];
}
printf("%d %d\n", len, ans);
}
return 0;
}