标签:operator opera 前缀 for query pre put using bit
因为所有数的和才100w,所以我们可以直接求出所有区间和。
直接把前缀和存到一个权值数组,再倒着存一遍,大力卷积一波。
这样做在bzoj目前还过不了,但是luogu开O2,最慢的点才500ms左右。
#include<bits/stdc++.h>
#define For(i,x,y) for (register int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (register int i=(x);i>=(y);i--)
#define cross(i,u) for (register int i=first[u];i;i=last[i])
using namespace std;
typedef long long ll;
inline ll read(){
ll x=0;int ch=getchar(),f=1;
while (!isdigit(ch)&&(ch!=‘-‘)&&(ch!=EOF)) ch=getchar();
if (ch==‘-‘){f=-1;ch=getchar();}
while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
return x*f;
}
const int N = 4e6+10;
int m,ans,pre[N];
struct node{
double x,y;
inline node operator + (const node &b)const{return (node){x+b.x,y+b.y};}
inline node operator - (const node &b)const{return (node){x-b.x,y-b.y};}
inline node operator * (const node &b)const{return (node){x*b.x-y*b.y,x*b.y+y*b.x};}
}a[N],b[N];
int n,l,pos[N];
inline void init(){
for (n=1;n<=(pre[m]<<1);n<<=1,l++);
For(i,0,n-1) pos[i]=(pos[i>>1]>>1)|((i&1)<<(l-1));
}
const double pi = 3.1415926535897932;
inline void FFT(node *a,int f){
For(i,0,n-1) if (i<pos[i]) swap(a[i],a[pos[i]]);
node x,y;
for (register int i=1;i<n;i<<=1){
node wn=(node){cos(pi/i),f*sin(pi/i)};
for (register int j=0;j<n;j+=i<<1){
node w=(node){1,0};
for (register int k=0;k<i;k++,w=w*wn) x=a[j+k],y=w*a[j+k+i],a[j+k]=x+y,a[j+k+i]=x-y;
}
}
if (f<0) For(i,0,n-1) a[i].x=a[i].x/n+0.5;
}
int main(){
m=read(),a[0].x=1;
For(i,1,m) pre[i]=pre[i-1]+read(),++a[pre[i]].x;
For(i,0,pre[m]) b[pre[m]-i].x=a[i].x;
init();FFT(a,1),FFT(b,1);
For(i,0,n-1) a[i]=a[i]*b[i];
FFT(a,-1);
//For(i,0,n-1) printf("%d ",(int)a[i].x);puts("");
For(i,0,pre[m]) if ((int)a[i].x&1) ans^=(-(i-pre[m]));
printf("%d",ans);
}
考虑拆位,大力分类讨论,再加一个树状数组维护下。
#include<bits/stdc++.h>
#define For(i,x,y) for (register int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (register int i=(x);i>=(y);i--)
#define cross(i,u) for (register int i=first[u];i;i=last[i])
using namespace std;
typedef long long ll;
inline ll read(){
ll x=0;int ch=getchar(),f=1;
while (!isdigit(ch)&&(ch!=‘-‘)&&(ch!=EOF)) ch=getchar();
if (ch==‘-‘){f=-1;ch=getchar();}
while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
return x*f;
}
const int N = 1e6+10;
int n,cnt,Max,ans,pre[N];
struct BIT{
int c[N];
inline void Add(int x,int y){for (;x<=pre[n];x+=x&-x) c[x]+=y;}
inline int Query(int x){int ans=0;for (;x;x-=x&-x) ans+=c[x];return ans;}
}t[2];
int main(){
n=read();
For(i,1,n) pre[i]=pre[i-1]+read();
for (int i=1;i<=pre[n];i<<=1){
cnt=Max=0;
For(j,0,n){
int x=((pre[j]&i)>0),y=pre[j]&(i-1);
Max=max(Max,y),cnt+=t[x^1].Query(y+1)+t[x].Query(Max+1)-t[x].Query(y+1),t[x].Add(y+1,1);
}
For(j,0,pre[n]) t[0].c[j]=t[1].c[j]=0;
ans|=(cnt&1)*i;
}
printf("%d",ans);
}
标签:operator opera 前缀 for query pre put using bit
原文地址:https://www.cnblogs.com/zykykyk/p/9502021.html