| Time Limit: 1000MS | Memory Limit: 30000K | |
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
题意：给出n个数，求最长单调递减子序列的长度L，并求出有多少个序列的长度为L。
注意：5 3 1， 5 3 1算一种方案。
分析：因为n不超过5000，所以O(n^2)复杂度可以接受。那么我们就可以按照平时求最长单调递减子序列的方法，在求的时候加上一个辅助数组cnt[i]记录从开始到第i个位置，单调递减子序列为dp[i]时的方案数。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e4 + 10;
int a[N], dp[N], cnt[N];
int main() {
int n;
while(~scanf("%d", &n)) {
for(int i = 0; i < n; i++) {
scanf("%d", &a[i]);
dp[i] = 1;
cnt[i] = 1;
}
for(int i = 1; i < n; i++) {
for(int j = i - 1; j >= 0; j--) {
if(a[i] < a[j]) {
if(dp[i] < dp[j] + 1) { // 第一次找到能使长度变长的元素
dp[i] = dp[j] + 1;
cnt[i] = cnt[j];
}
else if(dp[i] == dp[j] + 1) // 之前已经找到了一个使得dp[i] = dp[j] +１的元素，现在又找到一个
cnt[i] += cnt[j];
}
else {
if(a[i] == a[j]) {
if(dp[i] == 1) //用第i个位置上的数字替换第j个位置上的数字，形成的序列完全相同
cnt[i] = 0;
break;
}
}
}
}
int max_len = 0, ans_cnt = 0;
for(int i = 0; i < n; i++)
max_len = max(max_len, dp[i]);
for(int i = 0; i < n; i++)
if(dp[i] == max_len)
ans_cnt += cnt[i];
printf("%d %d\n", max_len, ans_cnt);
}
return 0;
}