原题链接
思路:
先观察数据范围,但是
,也就是说在
的数转化为二进制数后最多有
位,所以每次改变影响的只有
位。
可以先统计一遍第一次改变后的答案。对于后面的每次改变,假设当前改变的位置为,消除
的贡献,改变后加上
的贡献。
在统计答案的时候,枚举起点和位数,如果大于直接跳出即可。
还要注意不能含有前导零,枚举起点的时候遇到应该直接跳过
代码:
// Problem: 零一奇迹
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11214/G
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read(){ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}
inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
const int maxn=2e5+100,inf=0x3f3f3f3f;
const double eps=1e-5;
string s;
ll tr[maxn];
int n;
ll l,r;
bool check(string t){
ll ans=0;
for(int i=0;t[i];i++)
ans=ans*2+(t[i]-'0');
if(ans>=l&&ans<=r) return 1;
return 0;
}
ll cul(int L,int R){
ll ans=0;
for(int i=L;i<=R;i++){
if(s[i]=='0') continue;
ll tmp=0;
for(int j=1;j<=64&&i+j-1<=n;j++){
int now=i+j-1;
tmp=tmp*2+(s[now]-'0');
if(tmp>=l&&tmp<=r) ans++;
if(tmp>r) break;
}
}
return ans;
}
int main(){
#ifdef LOCAL
freopen("in.in","r",stdin);
freopen("out.out","w",stdout);
#endif
n=read;
l=read,r=read;
cin>>s;
s=" "+s;
int q=read;
ll ans=0;
for(int Case=1;Case<=q;Case++){
int x=read;
if(Case==1){
if(s[x]=='0') s[x]='1';
else s[x]='0';
ans=cul(1,n);
}
else{
ll t1=cul(max(1,x-63),x);
if(s[x]=='0') s[x]='1';
else s[x]='0';
ll t2=cul(max(1,x-63),x);
ans=ans-t1+t2;
}
printf("%lld\n",ans);
}
return 0;
}