Usaco buy low, buy Lower

This question has two questions, the first question is the longest descending subsequence.

For the second ask the number of the longest descending sequence can be solved by asking the first question process. Set Maxcnt[i] to be the number of the longest descending sequence at the end of item I.

For all J (1≤j≤i-1) if there is (S[j]>s[i] and maxlength[j]+1>maxlength[i]) then maxcnt[i]=maxcnt[j], otherwise if (maxlength[j]+1= = Maxlength[i]) can take advantage of the addition principle, maxcnt[i]=maxcnt[i]+maxcnt[j].

Taking into account the problem can not repeat the sequence, we can add a domain next[i] is greater than I and the nearest I next[i] so that the number of next[i] is the same as the number of first. If there is no such number then next[i]=0. This way we are in DP when the NEXT[J] is not 0 and Next[j]<i can skip directly.

/*id:modengd1prob:buylowlang:c++*/#include <iostream> #include <stdio.h> #include <memory.h># Include <vector>using namespace Std;int n;long input[5001];int dp[5001],next[5001];int c[5001][500];//    maxcnt array int main () {freopen ("buylow.in", "R", stdin);    Freopen ("Buylow.out", "w", stdout);    scanf ("%d", &n);    for (int i=0;i<n;i++) scanf ("%ld", &input[i]);    input[n]=0;    Memset (Dp,0,sizeof (DP));                for (int i=0;i<=n;i++) {for (int j=0;j<i;j++) {if (Input[j]>input[i]) {            Dp[i]=max (Dp[i],dp[j]);    }} dp[i]++;    } memset (Next,0,sizeof (next));                for (int i=0;i<=n;i++) {for (int j=i+1;j<=n;j++) {if (Input[i]==input[j]) {                Next[i]=j;            Break    }}} memset (C,0,sizeof (c)); for (int i=0;i<=n;i++) {for (int j=0;j<i;j++) {if (next[j]==0| | Next[j]>i) &&dp[j]==dp[i]-1&&input[j]>input[i]) {int k; for (k=1;k<=c[j][0];k++) {c[i][k]+=c[j][k];c[i][k+1]+=c[i][k]/10;c[i][k]%=10;} while (c[i][k]>=10) {c[i][k+1]+=c[i][k]/10;c[i][k]%=10;k++;}            while (c[i][k]==0) k--;if (k>=c[i][0]) c[i][0]=k;            }} if (c[i][0]==0) {c[i][0]=1;        C[i][1]=1;    }} cout<<dp[n]-1<< ';    for (int i=c[n][0];i>0;i--) cout<<c[n][i];    cout<<endl; return 0;}

　　

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Usaco buy low, buy Lower