将 \(1\) 的权值设为 \(1\),\(0\)的权值设为 \(-1\)。则一个子序列的分数为祂里面所有数的权值和除以 \(4\) 下取整。
把下取整去掉，变成 \(\frac{S_T^2-S_T\mod2}{4}\) 即 \(\frac{S_T^2-\lvert T\rvert\mod2}{4}\)
整个序列的分数为 \(\frac{\sum_T S_T^2-2^{n-1}}{4}\)
考虑求 \(\sum_T S_T^2\)
把平方展开 \(\sum_T(\lvert T\rvert+2\times\sum_{x,y\in T\cap x<y}val_x\times val_y)\) 即 \(n\times 2^{n-1}+2\times2^{n-2}\times\sum_{x<y}val_x\times val_y\)
\(\sum_{x<y}val_x\times val_y=\frac{(\sum val_x)^2-\sum val_x^2}{2}=\frac{(\sum val_x)^2-n}{2}\)
所以答案为
\(\frac{\frac{(\sum val_x)^2-n}{2}\times2^{n-2}+2\times2^{n-1}-s^{n-1}}{4}\) 即 \(\frac{2^{n-2}\times((\sum val_x)^2+n)-2^{n-1}}{4}\)
对于每个修改，维护 \((\sum val_x)^2\) 即可
代码
#include<bits/stdc++.h>
using namespace std;
namespace IO{template<typename T>inline void read(T&x){x=0;char c=getchar();bool f=0;while(!isdigit(c)) c=='-'?f=1:0,c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();f?x=-x:0;}template<typename T>inline void write(T x){if(x==0){putchar('0');return ;}x<0?x=-x,putchar('-'):0;short st[50],top=0;while(x) st[++top]=x%10,x/=10;while(top) putchar(st[top--]+'0');}inline void read(char&c){c=getchar();while(isspace(c)) c=getchar();}inline void write(char c){putchar(c);}inline void read(string&s){s.clear();char c;read(c);while(!isspace(c)&&~c) s+=c,c=getchar();}inline void write(string s){for(int i=0,len=s.size();i<len;i++) putchar(s[i]);}template<typename T>inline void write(T*x){while(*x) putchar(*(x++));}template<typename T,typename...T2> inline void read(T&x,T2&...y){read(x),read(y...);}template<typename T,typename...T2> inline void write(const T x,const T2...y){write(x),putchar(' '),write(y...),sizeof...(y)==1?putchar('\n'):0;}
}using namespace IO;
#define int long long
const int mod=998244353;
int n,q;
string s;
int ksm(int a,int b){int ans=1;b=(b+mod-1)%(mod-1);while(b){if(b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;}return ans;
}
void solve(){read(n,q,s);int sum=0;for(int i:s) if(i=='0') sum--;else sum++;for(int i=1;i<=q;i++){int w;read(w);if(s[w-1]=='0') sum+=2,s[w-1]='1';else sum-=2,s[w-1]='0';write((ksm(2,n-2)*(sum*sum%mod+n)%mod-ksm(2,n-1)+mod)%mod*ksm(4,mod-2)%mod);write('\n');}
}
signed main(){int T;read(T);while(T--) solve();return 0;
}