The romantic hero

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• Question:
  N number. Find two subscripts I and j (I <j). In 1-I, select the bitwise AND value of several numbers equal to or equal to J-n, the two sets are not empty. Evaluate the number of sets that meet the condition.
• Analysis:
  For an I instance, if you know all the values on the left and all the values on the right, the flight is part of the answer, and the DP is used to pre-process the values on one side.
  Focus on the DP status representation: DP [I] [J] indicates the number of cases where the number ends with an I-position number and the operation value is J, here, it indicates the sum of any number that can be combined and J. If it does not indicate that it ends with I, it will be repeated.
  In this case, why does Dp not repeat? For DP [x] [] and DP [y] [], because the maximum values of the two States are different, so the number of columns will not be the same, that is to say, the States represented by DP do not overlap. For a sequence, there is only one computing result, so it only belongs to a certain state of DP [x, it does not repeat (if a sequence can obtain multiple computing results, then DP cannot be like this, because a definite sequence can belong to multiple States ).

const int MAXN = 1024;int ipt[MAXN];LL dp1[MAXN][MAXN], dp2[MAXN][MAXN], sum1[MAXN][MAXN], sum2[MAXN][MAXN];int main(){    int T, n;    RI(T);    FE(kase, 1, T)    {        CLR(dp1, 0); CLR(sum1, 0);        CLR(dp2, 0); CLR(sum2, 0);        RI(n);        REP(i, n)            RI(ipt[i]);        dp1[0][ipt[0]] = sum1[0][ipt[0]] = 1;        FF(i, 1, n)        {            REP(j, MAXN)                dp1[i][j ^ ipt[i]] = (sum1[i - 1][j] + dp1[i][j ^ ipt[i]]) % MOD;            dp1[i][ipt[i]]++;            REP(j, MAXN)                sum1[i][j] = (sum1[i - 1][j] + dp1[i][j]) % MOD;        }        dp2[n - 1][ipt[n - 1]] = sum2[n - 1][ipt[n - 1]] = 1;        FED(i, n - 2, 0)        {            REP(j, MAXN)                dp2[i][j & ipt[i]] = (sum2[i + 1][j] + dp2[i][j & ipt[i]]) % MOD;            dp2[i][ipt[i]]++;            REP(j, MAXN)                sum2[i][j] = (sum2[i + 1][j] + dp2[i][j]) % MOD;        }        LL ans = 0;        REP(i, n - 1)            REP(j, MAXN)                ans = (ans + dp1[i][j] * sum2[i + 1][j]) % MOD;        cout << ans << endl;    }    return 0;}