Ultraviolet A 1406-A Sequence of Numbers (tree array)

Ultraviolet A 1406-A Sequence of Numbers (tree array)

Link to the question: A 1406-A Sequence of Numbers

There are two operations for a given number of n:

• Q x: calculates the number of values equal to or greater than zero for 2x.
• C x: Add x to each number.
  Outputs the sum of all Q operations.

  Solution: Because x is 15 at the maximum, 16 tree arrays are opened. fenx [x] records the situation where every number modulo is 2x + 1, then there is an add value to mark the total amount added. Determine the range of the original number based on the add value.

  #include 
      
       #include 
       
        #include using namespace std;#define lowbit(x) ((x)&(-x))const int maxn = 1<<16;const int maxr = 16;int base[maxr+5], fenx[maxr+5][maxn+5];int N, add;int query_treeArr (int* bit, int x) {    int ret = 0;    while (x) {        ret += bit[x];        x -= lowbit(x);    }    return ret;}void insert_treeArr (int* bit, int x, int v) {    while (x <= maxn) {        bit[x] += v;        x += lowbit(x);    }}void init () {    add = 0;    for (int i = 0; i < maxr; i++)         memset(fenx, 0, sizeof(fenx));    int x;    for (int i = 0; i < N; i++) {        scanf("%d", &x);        for (int j = 1; j <= maxr; j++) {            insert_treeArr(fenx[j-1], x % base[j] + 1, 1);        }    }}long long solve () {    long long ret = 0;    int x;    char order[5];    while (scanf("%s", order) == 1 && strcmp(order, "E")) {        scanf("%d", &x);        if (order[0] == 'Q') {            int have = add % base[x];            if (add & base[x]) {                ret += query_treeArr(fenx[x], base[x] - have);                ret += query_treeArr(fenx[x], base[x+1]) - query_treeArr(fenx[x], base[x+1] - have);            } else                ret += query_treeArr(fenx[x], base[x+1] - have) - query_treeArr(fenx[x], base[x] - have);        } else            add = (add + x) % base[16];    }    return ret;}int main () {    int cas = 1;    base[0] = 1;    for (int i = 1; i <= maxr; i++)        base[i] = base[i-1] * 2;    while (scanf("%d", &N) == 1 && N != -1) {        init();        printf("Case %d: %lld\n", cas++, solve());    }    return 0;}