fzu 2105(区间更新+位拆分)

题意：先给出n个数字，然后有q次询问，四种操作AND opn l r表示区间[l,r]里的数字和opn按位做&操作，OR opn l r表示区间[l,r]里的数字和opn按位做|操作，XOR opn l r表示区间[l,r]里的数字和opn按位做^操作，SUM l r表示输出区间[l,r]里所有数字的和。
题解：opn和A[i]的范围是15可以想到二进制只有4位，那么可以想到数位分离，把每一位单独拿出来考虑，那就可以把问题转化为区间内所有数字在某一位全部变为0(AND操作)或变为1(OR操作)或01数目互换(XOR操作)，所以建4个线段树维护区间内某个位的0和1的个数，最后按二进制转十进制方法计算总和，这里有两个标记传递，一个是赋值另一个是置换，如果先置换再赋值会被覆盖结果，所以要先传递覆盖再传递置换，同时在传递覆盖时把左右子区间的置换标记清除。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1000005;int n, q, a[N], bit, tree0[N << 2], tree1[N << 2], res[N][4];int flag1[N << 2], flag2[N << 2], l[N], r[N], v[N];char op[N][5];void pushup(int k) {    tree0[k] = tree0[k * 2] + tree0[k * 2 + 1];    tree1[k] = tree1[k * 2] + tree1[k * 2 + 1];}void pushdown(int k, int left, int right) {    if (flag1[k] != -1) {        int mid = (left + right) / 2;        flag1[k * 2] = flag1[k * 2 + 1] = flag1[k];        if (flag1[k] == 0) {            tree0[k * 2] = mid - left + 1;            tree0[k * 2 + 1] = right - mid;            tree1[k * 2] = tree1[k * 2 + 1] = 0;        }        else if (flag1[k] == 1) {            tree1[k * 2] = mid - left + 1;            tree1[k * 2 + 1] = right - mid;            tree0[k * 2] = tree0[k * 2 + 1] = 0;        }        flag1[k] = -1;        flag2[k * 2] = flag2[k * 2 + 1] = 0;    }    if (flag2[k]) {        flag2[k * 2] ^= flag2[k];        flag2[k * 2 + 1] ^= flag2[k];        swap(tree0[k * 2], tree1[k * 2]);        swap(tree0[k * 2 + 1], tree1[k * 2 + 1]);        flag2[k] = 0;    }}void build(int k, int left, int right, int bit) {    tree0[k] = tree1[k] = flag2[k] = 0;    flag1[k] = -1;    if (left == right) {        if ((a[left] & (1 << bit)))            tree1[k] = 1;        else tree0[k] = 1;        return;    }    int mid = (left + right) / 2;    build(k * 2, left, mid, bit);    build(k * 2 + 1, mid + 1, right, bit);    pushup(k);}void modify(int k, int left, int right, int l1, int r1, int flag_1, int flag_2) {    if (l1 <= left && right <= r1) {        if (flag_1 == 0) {            flag1[k] = flag_1;            tree0[k] = right - left + 1;            tree1[k] = flag2[k] = 0;        }        else if (flag_1 == 1) {            flag1[k] = flag_1;            tree1[k] = right - left + 1;            tree0[k] = flag2[k] = 0;        }        else {            flag2[k] ^= flag_2;            swap(tree0[k], tree1[k]);        }        return;    }    pushdown(k, left, right);    int mid = (left + right) / 2;    if (l1 <= mid)        modify(k * 2, left, mid, l1, r1, flag_1, flag_2);    if (r1 > mid)        modify(k * 2 + 1, mid + 1, right, l1, r1, flag_1, flag_2);    pushup(k);}int query(int k, int left, int right, int l1, int r1) {    if (l1 <= left && right <= r1)        return tree1[k];    pushdown(k, left, right);    int mid = (left + right) / 2;    if (r1 <= mid)        return query(k * 2, left, mid, l1, r1);    if (l1 > mid)        return query(k * 2 + 1, mid + 1, right, l1, r1);    return query(k * 2, left, mid, l1, mid) + query(k * 2 + 1, mid + 1, right, mid + 1, r1);}int main() {    int t;    scanf("%d", &t);    while (t--) {        scanf("%d%d", &n, &q);        for (int i = 1; i <= n; i++)            scanf("%d", &a[i]);        for (int i = 1; i <= q; i++) {            scanf("%s", op[i]);            if (op[i][0] == 'S')                scanf("%d%d", &l[i], &r[i]);            else                scanf("%d%d%d", &v[i], &l[i], &r[i]);        }        for (int i = 0; i < 4; i++) {            build(1, 1, n, i);            for (int j = 1; j <= q; j++) {                if (op[j][0] == 'S')                    res[j][i] = query(1, 1, n, l[j] + 1, r[j] + 1);                else if (op[j][0] == 'A' && !(v[j] & (1 << i)))                    modify(1, 1, n, l[j] + 1, r[j] + 1, 0, 0);                else if (op[j][0] == 'O' && (v[j] & (1 << i)))                    modify(1, 1, n, l[j] + 1, r[j] + 1, 1, 0);                else if (op[j][0] == 'X' && (v[j] & (1 << i)))                    modify(1, 1, n, l[j] + 1, r[j] + 1, -1, 1);            }        }        for (int i = 1; i <= q; i++)            if (op[i][0] == 'S')                printf("%d\n", res[i][0] + res[i][1] * 2 + res[i][2] * 4 + res[i][3] * 8);    }    return 0;}