题目:
题解:
暴力一波 (SG) 函数可以发现这么一个规律:
- (p) 为奇数的时候 : (SG(n) = n \% 2)
- (p) 为偶数的时候 : (SG(n) = n \% (p+1) == p ? 2 : n \% (p+1) \% 2)
对于奇数的情况我们就可以直接用一棵支持区间取反和区间查询 $1$ 的个数的线段树搞定。 那么难点在于偶数的情况。
我们可以采用分块算法. 每个块分别中保存 (mod (p+1)) 为奇数的数和为偶数的数。 然后每次查询的时候我们可以通过几次 (lower_bound) 来解决。 复杂度 (O(mlog n)space orspace O(msqrt{n}log n))
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int*,int*> pa;
inline void read(int &x){
x=0;static char ch;static bool flag;flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
#define rg register int
#define rep(i,a,b) for(rg i=(a);i<=(b);++i)
#define per(i,a,b) for(rg i=(a);i>=(b);--i)
const int maxn = 100010;
int n,q,p;
namespace work1{
int T[maxn<<2],tag[maxn<<2],a[maxn];
inline void push_down(int rt,int l,int r){
if(l == r || tag[rt] == 0) return ;
int mid = l+r >> 1;
tag[rt<<1] ^= 1;
if((mid-l+1)&1) T[rt<<1] ^= 1;
tag[rt<<1|1] ^= 1;
if((r-mid)&1) T[rt<<1|1] ^= 1;
tag[rt] = 0;
}
void build(int rt,int l,int r){
if(l == r){
T[rt] = a[l]&1;
return ;
}int mid = l+r >> 1;
build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);
T[rt] = T[rt<<1]+T[rt<<1|1] & 1;
}
int L,R;
void modify(int rt,int l,int r){
if(L <= l && r <= R){
tag[rt] ^= 1;
if((r-l+1)&1) T[rt] ^= 1;
return ;
}int mid = l+r >> 1;push_down(rt,l,r);
if(L <= mid) modify(rt<<1,l,mid);
if(R > mid) modify(rt<<1|1,mid+1,r);
T[rt] = T[rt<<1]+T[rt<<1|1] & 1;
}
int query(int rt,int l,int r){
if(L <= l && r <= R) return T[rt];
int mid = l+r >> 1;push_down(rt,l,r);
if(R <= mid) return query(rt<<1,l,mid);
if(L > mid) return query(rt<<1|1,mid+1,r);
return query(rt<<1,l,mid) + query(rt<<1|1,mid+1,r) & 1;
}
int main(){
rep(i,1,n) read(a[i]);
build(1,1,n);
int op,x;
while(q--){
read(op);
if(op == 0){
read(L);read(R);read(x);
if(x & 1) modify(1,1,n);
}else{
read(L);read(R);
printf("%d
",query(1,1,n));
}
}
return 0;
}
}
namespace work2{
const int maxm = 322;
int tmp[maxn],c[maxn];
struct Node{
int a[maxm],b[maxm],tag;
int cnta,cntb,l,r;
// a[i] % (p+1) = 1;
// b[i] % (p+1) = 0;
void build(int n){
sort(tmp+1,tmp+n+1);
cnta = cntb = 0;
rep(i,1,n){
if(tmp[i] % (p+1) & 1) a[++cnta] = tmp[i];
else b[++cntb] = tmp[i];
}return ;
}
int query_all(int &one,int &two){
int x = (p - tag + (p+1)) % (p+1);pa pos;
if(x & 1) pos = equal_range(a+1,a+cnta+1,x);
else pos = equal_range(b+1,b+cntb+1,x);
two += (pos.second - pos.first);
if(tag <= (p-1)){
x = p - 1 - tag;int ps = 0;
if(x & 1) ps = upper_bound(a+1,a+cnta+1,x) - a - 1;
else ps = upper_bound(b+1,b+cntb+1,x) - b - 1;
one += ps;
}
if(tag >= 2){
x = p - tag;int ps = 0;
if(x & 1) ps = cnta - (upper_bound(a+1,a+cnta+1,x)-a-1);
else ps = cntb - (upper_bound(b+1,b+cntb+1,x)-b-1);
one += ps;
}
}
void query(int l,int r,int &one,int &two){
int cnt = 0;
rep(i,this->l,this->r){
c[i] += tag;
if(c[i] >= p+1) c[i] -= p+1;
tmp[++cnt] = c[i];
}sort(tmp+1,tmp+cnt+1);
cnta = cntb = tag = 0;
rep(i,1,cnt){
if(tmp[i] % (p+1) & 1) a[++cnta] = tmp[i];
else b[++cntb] = tmp[i];
}
rep(i,l,r){
if(c[i] % (p+1) == p) ++ two;
else if(c[i] % (p+1) & 1) ++ one;
}return ;
}
void modify(int l,int r,int x){
int cnt = 0;
rep(i,this->l,this->r){
c[i] += tag;if(c[i] >= (p+1)) c[i] -= (p+1);
if(l <= i && i <= r){
c[i] += x;
if(c[i] >= (p+1)) c[i] -= (p+1);
}
tmp[++cnt] = c[i];
}sort(tmp+1,tmp+cnt+1);
cnta = cntb = tag = 0;
rep(i,1,cnt){
if(tmp[i] % (p+1) & 1) a[++cnta] = tmp[i];
else b[++cntb] = tmp[i];
}return ;
}
}zs[maxm];
int belong[maxn];
inline void modify(int l,int r,int x){
if(belong[l] == belong[r]) return zs[belong[l]].modify(l,r,x);
zs[belong[l]].modify(l,zs[belong[l]].r,x);
zs[belong[r]].modify(zs[belong[r]].l,r,x);
rep(i,belong[l]+1,belong[r]-1){
zs[i].tag += x;
if(zs[i].tag >= (p+1)) zs[i].tag -= (p+1);
}
return ;
}
inline int query(int l,int r){
int res1 = 0,res2 = 0;
if(belong[l] == belong[r]){
zs[belong[l]].query(l,r,res1,res2);
}else{
zs[belong[l]].query(l,zs[belong[l]].r,res1,res2);
zs[belong[r]].query(zs[belong[r]].l,r,res1,res2);
rep(i,belong[l]+1,belong[r]-1) zs[i].query_all(res1,res2);
}
if((res1 & 1) || (res2 & 1)) return 1;
else return 0;
}
int main(){
int block = ceil(sqrt(n));
int sz = 0,m = 1;
zs[1].l = 1;
rep(i,1,n){
belong[i] = m;read(tmp[++sz]);
c[i] = (tmp[sz] %= (p+1));
if(sz == block){
zs[m].build(sz);
zs[m].r = i,zs[++m].l = i+1;
sz = 0;
}
}
if(zs[m].l != n+1){
zs[m].r = n;
zs[m].build(n-zs[m].l+1);
}else --m;
int op,l,r,x;
while(q--){
read(op);read(l);read(r);
if(op == 0) read(x),modify(l,r,x % (p+1));
else printf("%d
",query(l,r));
}
return 0;
}
}
int main(){
read(n);read(q);read(p);
if(p & 1) work1::main();
else work2::main();
return 0;
}