题目链接：uva 1406 - A Sequence of Numbers

题目大意；给定n个数，有两种操作：

• Q x：计算与2x取且不为0的数的个数
• C x：每个数加上x
  输出所有Q操作的和。

解题思路：因为x最大为15，所以开16个树状数组，fenx[x]记录的是每个数取模2x+1的情况，然后有一个add值标记总共加了多少。根据add值确定原先数的范围。

#include <cstdio>
#include <cstring>
#include <algorithm>

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;
}

uva 1406 - A Sequence of Numbers(树状数组)