poj 1952 BUY LOW, BUY LOWER

dp:
求一串序列中最長遞減子序列包含的數的個數以及有幾個長度相同的最長的遞減子序列，並不要求連續。

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
const int size = 5100;
int maxlen[size];
int maxnum[size];
int main()
{
    int a[size];
    int n;
    while (scanf("%d", &n) != EOF){
          for (int i = 1; i <= n; i ++){
              scanf("%d", &a[i]);
              maxnum[i] = 0;
              maxlen[i] = 1;
          }
          for (int i = 1 ; i <= n; i ++){
              for (int j = 1; j < i; j ++){
                  if (a[i] < a[j]){
                     maxlen[i] = max(maxlen[i], maxlen[j]+1);//我們只要比較每一個a[i]和它前面的a[j](j<i),如果a[i]>a[j],則到i爲止的的最長序列至少比到j多1.
                  }
              }
          }
          for (int i = 1; i <= n; i ++)
              if (maxlen[i] == 1)maxnum[i] = 1;
          for (int i = 1; i <= n; i ++){
              for (int j = i-1; j > 0; j --){
                  if (a[j] > a[i]){
                     if (maxlen[j]+1 == maxlen[i]){
                        maxnum[i] += maxnum[j];
                     }
                  }
                  if (a[j] == a[i]){
                     if (maxlen[i] == 1)maxnum[i] = 0;
                     break;
                  }
              }
          }
          int maxx = -1;
          for (int i = 1; i <= n; i ++){
              if (maxlen[i] > maxx)  maxx = maxlen[i];
          }
          int ans = 0;
          for (int i = 1; i <= n; i ++){
              if (maxlen[i] == maxx) ans += maxnum[i] ;
          }
          printf("%d %d\n", maxx, ans);
    }
    return 0;
}