HDU 4455 substrings (DP)

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Ideas:

We use d[i] to denote the answer of length I.   Then we can write the sample data to see if we can push it. Can be found, d[i] = d[i-1]-last[i-1] + (n-i+1)-dist[i].

Last[i] Indicates the number of different numbers from the backward I length, that is, the last subsequence of D[i-1] is thrown, and then d[i] may be added (N-i + 1) than d[i-1], and then we subtract the non-conforming requirements. When the competition was launched, it is not clear, suggest to push yourself.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> Using namespace Std;typedef long Long ll;const double PI = ACOs ( -1.0); const double EPS = 1e-6;const int INF = 1000000000;c    onst int maxn = 1e6+5;int T,N,M,R,A[MAXN],LAST[MAXN],DIST[MAXN], q, x, P[maxn];ll d[maxn];bool hehe[maxn];int main () {        while (~SCANF ("%d", &n) && N) {memset (d, 0, sizeof (d));        memset (hehe, 0, sizeof (hehe));        memset (Dist, 0, sizeof (Dist));        D[1] = n;        for (int i = 1; I <= n; i++) {scanf ("%d", &a[i]);        } int cnt = 1;        memset (p, 0, sizeof (p));            for (int i = n; I >= 1; i--) {if (P[a[i]]) last[cnt] = last[cnt-1]; else {P[a[i]] = 1;            LAST[CNT] = last[cnt-1] + 1;        } cnt++;        } memset (p, 0, sizeof (p));                for (int i = 1; I <= n; i++) {if (P[a[i]]) {dist[i-p[a[i]]+1]++;                Hehe[i] = 1;            P[a[i]] = i;        } else p[a[i]] = i;            } for (int i = 1; I <= n; i++) {if (hehe[i-1]) dist[i] + = dist[i-1]-1;        else Dist[i] + = dist[i-1];        } for (int i = 2; I <= n; i++) {d[i] = D[i-1]-(LL) Last[i-1] + (LL) (N-i + 1)-(ll) dist[i];        } scanf ("%d", &q);            while (q--) {scanf ("%d", &x);        printf ("%i64d\n", d[x]); }} return 0;}

HDU 4455 substrings (DP)