树状数组维护前缀和、后缀和、个数|牛牛的Link Power

思路

另线段树做法：https://www.cnblogs.com/fisherss/p/12287606.html

F题 2棵树状数组维护前缀和和个数即可

题目地址
https://ac.nowcoder.com/acm/contest/3004/F

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1e5+100;
int n;
char s[maxn];
const ll mod = 1e9+7;

struct bit{
	ll c[maxn];
	
	int lowbit(int x){
		return x & -x;
	}
	
	void add(int pos,int v){
		while(pos <= n){
			c[pos] += v;
			pos += lowbit(pos);
		}
	}
	
	ll getsum(int pos){
		ll res = 0;
		while(pos){
			res += c[pos];
			pos -= lowbit(pos);
		}
		return res;
	}
	
}pre,suf,cnt;

int main(){
	cin>>n;
	scanf("%s",s+1);
	ll ans = 0;
	for(int i=1;i<=n;i++){
		if(s[i] == '1'){
			cnt.add(i,1);
			pre.add(i,i);
			ans = (ans + i*cnt.getsum(i) - pre.getsum(i))%mod; 
		}
	}
	cout<<ans<<endl;
	return 0;
} 

G题 3棵树状数组维护前缀和、后缀和、个数

题目地址
https://ac.nowcoder.com/acm/contest/3004/G

70%,wa

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1e5+100;
ll n;
char s[maxn];
const ll mod = 1e9+7;

struct bit{
	ll c[maxn];
	
	ll lowbit(ll x){
		return x & -x;
	}
	
	void add(ll pos,ll v){
		while(pos <= n){
			c[pos] += v;
			pos += lowbit(pos);
		}
	}
	
	ll getsum(ll pos){
		ll res = 0;
		while(pos){
			res += c[pos];
			pos -= lowbit(pos);
		}
		return res;
	}
	
}pre,suf,cnt;

void print(){
	for(int i=1;i<=n;i++){
		cout<<suf.getsum(i)<<" ";
	}
	cout<<endl;
}

int main(){
	cin>>n;
	scanf("%s",s+1);
	ll ans = 0;
	for(ll i=1;i<=n;i++){
		if(s[i] == '1'){
			cnt.add(i,1);
			pre.add(i,i);
			suf.add(i,n-i+1);
			ans = (ans + i*cnt.getsum(i) - pre.getsum(i))%mod; 
		}
	}
	cout<<ans%mod<<endl;
	ll m;
	cin>>m;
	while(m--){
		ll q,pos;
		cin>>q>>pos;
		if(q == 1){
			//加点 
			cnt.add(pos,1);
			pre.add(pos,pos);
			suf.add(pos,n-pos+1);
			//加上对前缀1的贡献影响 
			ans = (ans + (pos*cnt.getsum(pos) - pre.getsum(pos)))%mod;
			//加上对后缀1的贡献影响 
			ans = (ans + ((n - pos + 1) * (cnt.getsum(n) - cnt.getsum(pos-1)) - (suf.getsum(n) - suf.getsum(pos-1))))%mod;
		}else{
			ans = (ans - (pos*cnt.getsum(pos) - pre.getsum(pos)))%mod;
			ans = (ans - ((n - pos + 1) * (cnt.getsum(n) - cnt.getsum(pos-1)) - (suf.getsum(n) - suf.getsum(pos-1))))%mod;
			cnt.add(pos,-1);
			pre.add(pos,-pos);
			suf.add(pos,-(n-pos+1));
		}
		cout<<ans%mod<<endl;
	}
	return 0;
}