Tree-like array + bitwise operations LA 4013 A Sequence of Numbers

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Test instructions: N number, two operations, one is each number plus X, second is the number of query & (1 << T) = = 1

Analysis: Because the accumulation is always, so can be processed offline. The tree array point is c[16][m] that represents the number X (1 << J) after the digital POS, considering the number of the first J bits. When asked, depending on the value of the add different processing conditions.

#include <bits/stdc++.h>using namespace Std;typedef long long ll;const int N = 1e5 + 5;const int R = (int) 1 <&lt ;    16;const int M = R + 10;struct BIT {int c[16][m];    void init (void) {memset (c, 0, sizeof (c));  } void Updata (int b, int pos) {pos++;            POS may be equal to 0 while (Pos < M) {C[b][pos] + = 1;        POS + pos &-pos;        }} int sum (int b, int pos) {pos++;        int ret = 0;            while (pos > 0) {ret + = C[b][pos];        pos-= pos &-pos;    } return ret;    }}bit;int Main (void) {int n, cas = 0;        while (scanf ("%d", &n) = = 1) {if (n = =-1) break;        Bit.init ();            for (int x, i=0; i<n; ++i) {scanf ("%d", &x);            for (int j=0; j<16; ++j) {Bit.updata (j, x% (1 << (j + 1))); }} ll add = 0, ans = 0;        Char str[2]; while (scanf ("%s", &str) = = 1{if (str[0] = = ' E ') break;  if (str[0] = = ' C ') {//offline int x;                scanf ("%d", &x);                add + = x;            if (add >= R) Add%= R;  } else {int t;                scanf ("%d", &t);                int tail = add% (1 << t); if (Add & (1 << t)) {//(1&LT;&LT;T) bits already have 1 ans + = bit.sum (t, (1 << T)-1-        tail); After +tail, both are 0 ans + = bit.sum (t, (1 << (T + 1))-1)-Bit.sum (T, (1 << (T + 1))-1-    tail);  +tail before 1, 0} else {ans + bit.sum (t, (1 << (T + 1))-1-   Tail)-Bit.sum (T, (1 << T)-1-tail);    +tail 1}}} printf ("Case%d:%lld\n", ++cas, ans); } return 0;}

　　

Tree-like array + bitwise operations LA 4013 A Sequence of Numbers