CF1734F Zeros and Ones

Description

https://www.luogu.com.cn/problem/CF1734F

Solution

观察到 $S_i=\text{popcount}(i)\bmod 2$。

所以我们要求的就是 $\sum\limits_{i=0}^{m-1} \text{popcount}(i\bigoplus(i+n))\bmod 2$。

考虑在二进制上,维护 $i$ 和 $i+n$ 的加法,只需要记录位数,是否向后面进位,目前 $\text{popcount}$ 的奇偶。就可以数位 dp 了。

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#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define maxn
#define put() putchar('\n')
#define Tp template<typename T>
#define Ts template<typename T,typename... Ar>
using namespace std;
Tp void read(T &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
namespace Debug{
Tp void _debug(char* f,T t){cerr<<f<<'='<<t<<endl;}
Ts void _debug(char* f,T x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
#define fi first
#define se second
#define mk make_pair
const int mod=1e9+7;
int power(int x,int y=mod-2) {
int sum=1;
while (y) {
if (y&1) sum=sum*x%mod;
x=x*x%mod;y>>=1;
}
return sum;
}
int n,m;
int a[65],na,b[65],nb;
int f[65][2][2][2];
int dfs(int id,int pos,int fl,int lim) {
if (id==0) return pos==0&&fl==0;
if (f[id][pos][fl][lim]!=-1) return f[id][pos][fl][lim];
int res=0;
int r=(lim?a[id]:1);
int i,j,k;
for (i=0;i<=r;i++) {
int tmp=i+b[id]-pos*2;
if (tmp>=-1&&tmp<=0) res+=dfs(id-1,1,fl^i^(tmp+1),lim&&(i==r));
if (tmp>=0&&tmp<=1) res+=dfs(id-1,0,fl^i^tmp,lim&&(i==r));
}
return f[id][pos][fl][lim]=res;
}
void solve(void) {
int i,now;
read(n);read(m);na=nb=0;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(f,-1,sizeof(f));
m--;now=m;
while (now) a[++na]=now%2,now/=2;
now=n;
while (now) b[++nb]=now%2,now/=2;
printf("%lld\n",dfs(max(na,nb)+1,0,1,1));
}
signed main(void){
int T;
read(T);
while (T--) solve();
return 0;
}