codeforces 242E(线段树)

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题意：两个操作：1,查询区间和 2.对区间的所有的值进行异或x;
思路：线段数维护一个区间的每一位的1的个数。
#include<bits/stdc++.h>using namespace std;#define clr(x,y) memset(x,y,sizeof x)typedef long long ll;const int maxn = 2e5 + 10;int a[maxn];int add[maxn];struct Node{    Node(){clr(dig,0);}    int dig[22];}tree[maxn << 2];void build(int l,int r,int rt){    if(l == r)    {        for(int i = 0; i < 22;i ++)        {            if(a[l] & (1 << i))tree[rt].dig[i] ++;        }        return ;    }    int mid = (l + r) >> 1;    build(l,mid,rt << 1);build(mid + 1,r,rt << 1|1);    for(int i = 0; i < 22;i ++)    {        tree[rt].dig[i] = tree[rt << 1].dig[i] + tree[rt << 1|1].dig[i];    }}void pushdown(int rt,int l,int r){    int mid = (l + r) >> 1;    if(add[rt] == 0)return;    add[rt << 1] ^= add[rt];add[rt << 1|1] ^= add[rt];    for(int i = 0; i < 22;i ++)    {        if(add[rt] & (1 << i))        {            tree[rt << 1].dig[i] = mid - l + 1 - tree[rt << 1].dig[i];            tree[rt << 1 | 1].dig[i] = r - mid - tree[rt << 1 |1].dig[i];                    }    }    add[rt] = 0;}void update(int L,int R,int x,int l,int r,int rt){    if(L <= l && R >= r)    {        for(int i = 0;i < 22;i ++)        {            if(x & (1 << i))            {                tree[rt].dig[i] = r - l + 1 - tree[rt].dig[i];            }        }        add[rt] ^= x;        return ;    }    pushdown(rt,l,r);    int mid = (l + r) >> 1;    if(L <= mid)update(L,R,x,l,mid,rt << 1);    if(R >= mid + 1)update(L,R,x,mid + 1,r,rt << 1|1);    for(int i = 0; i < 22;i ++)    {        tree[rt].dig[i] = tree[rt << 1].dig[i] + tree[rt << 1|1].dig[i];    }}Node query(int L,int R,int l,int r,int rt){    if(L <= l && R >= r)    {        return tree[rt];    }    pushdown(rt,l,r);    int mid = (l + r) >> 1;    Node t1,t2,t;    if(L <= mid)t1 = query(L,R,l,mid,rt << 1);    if(R >= mid + 1)t2 = query(L,R,mid + 1,r,rt << 1|1);    for(int i = 0;i < 22;i ++)    t.dig[i] = t1.dig[i] + t2.dig[i];    return t;}int main(){    // freopen("in.txt","r",stdin);    int n;    while( ~ scanf("%d",&n))    {        clr(add,0);        for(int i = 1; i < maxn << 2;i ++)clr(tree[i].dig,0);        for(int i = 1;i <= n;i ++)scanf("%d",&a[i]);        build(1,n,1);        int m;scanf("%d",&m);        while(m --)        {            int type;scanf("%d",&type);            if(type == 1)            {                int l,r;scanf("%d%d",&l,&r);                Node t = query(l,r,1,n,1);                ll ans = 0;                for(int i = 0; i < 22;i ++)                ans += (1ll << i) * t.dig[i];                printf("%lld\n",ans);            }            else             {                int l,r,x;scanf("%d%d%d",&l,&r,&x);                update(l,r,x,1,n,1);            }        }    }    return 0;}
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