题目大意

有一个二进制数，让你进行以下操作：

1. 将一个区间的数字按升/降序排列
2. 查询一个区间的数字构成的数

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解题思路

可以用线段树来维护每一位，预处理出2的整次幂，上传时把左儿子乘上右儿子长度次幂即可

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code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 200210
#define wyc 1000000007
using namespace std;
ll n,m,x,y,z,a[N],v[N];
char s[N];
struct node
{node(ll lenn=0,ll numm=0,ll summ=0){len=lenn;num=numm;sum=summ;}ll len,num,sum;
};
node merge(node a,node b)
{node c;c.len=a.len+b.len;c.num=(a.num*v[b.len]%wyc+b.num)%wyc;c.sum=a.sum+b.sum;return c;
}
struct Tree
{#define ls x*2#define rs x*2+1node s[N<<2];ll lazy[N<<2];void push_up(ll x){s[x]=merge(s[ls],s[rs]);return;}void get(ll x,ll y){lazy[x]=y;s[x].num=(v[s[x].len]+wyc-1)%wyc*y;s[x].sum=s[x].len*y;return;}void push_down(ll x){if(lazy[x]!=-1){get(ls,lazy[x]);get(rs,lazy[x]);lazy[x]=-1;}return;}void build(ll x,ll l,ll r){lazy[x]=-1;if(l==r){s[x]=node(1,a[l],a[l]);return;}ll mid=l+r>>1;build(ls,l,mid);build(rs,mid+1,r);push_up(x);return;}void change(ll x,ll L,ll R,ll l,ll r,ll y){if(L==l&&R==r){get(x,y);return;}ll mid=L+R>>1;push_down(x);if(r<=mid)change(ls,L,mid,l,r,y);else if(l>mid)change(rs,mid+1,R,l,r,y);else change(ls,L,mid,l,mid,y),change(rs,mid+1,R,mid+1,r,y);push_up(x);return;}node ask(ll x,ll L,ll R,ll l,ll r){if (L==l&&R==r)return s[x];ll mid=L+R>>1;push_down(x);if(r<=mid)return ask(ls,L,mid,l,r);else if(l>mid)return ask(rs,mid+1,R,l,r);else return merge(ask(ls,L,mid,l,mid),ask(rs,mid+1,R,mid+1,r));}
}T;
int main()
{scanf("%s",s+1);n=strlen(s+1);v[0]=1;for(ll i=1;i<=n;++i){a[i]=s[i]-48;v[i]=v[i-1]*2%wyc;}T.build(1,1,n);scanf("%lld",&m);while(m--){scanf("%lld",&x);if(x==1){scanf("%lld%lld",&x,&y);z=T.ask(1,1,n,x,y).sum;if(!z||z==y-x+1)continue;T.change(1,1,n,x,y,0);T.change(1,1,n,x,x+z-1,1);}else if(x==2){scanf("%lld%lld",&x,&y);z=T.ask(1,1,n,x,y).sum;if(!z||z==y-x+1)continue;T.change(1,1,n,x,y,0);T.change(1,1,n,y-z+1,y,1);}else{scanf("%lld%lld",&x,&y);printf("%lld\n",T.ask(1,1,n,x,y).num%wyc);}}return 0;
}