2020牛客寒假算法基础集训营3 G.牛牛的Link Power II （树状数组维护前缀和）

https://ac.nowcoder.com/acm/contest/3004/G

发现每个“1”对于它本身位置产生的影响贡献为0，对前面的“1”有产生贡献，对后面的"1"也产生贡献，用三个树状数组去维护，第一个维护“1”的个数，第二个每个“1”的位置，第三个维护“1”的贡献的前缀和，对于每个“1”的贡献，计算出对前面1的贡献和对后面1的贡献，之和在第一个树状数组中更新这个位置1的个数，第二个树状数组中更新这个1的pos，第三个树状数组中就是更新这个“1”对答案的贡献了

 

 #include<bits/stdc++.h>
 typedef long long ll;
 using namespace std;
 const ll mod = 1e9+;
 const int maxn = 1e5+;
 ll c1[maxn],c2[maxn],c3[maxn];
 int n;
 ll lowbit(ll x){
   return x&-x;
 }
 void add1(ll x, ll k) {
   while (x <= n) {  //不能越界
     c1[x] = c1[x] + k;
     x = x + lowbit(x);
   }
 }
 void add2(ll x, ll k) {
   while (x <= n) {  //不能越界
     c2[x] = (c2[x] + k);
     x = x + lowbit(x);
   }
 }
 void add3(ll x,ll k){
   while (x <= n) {  //不能越界
     c3[x] = (c3[x] + k);
     x = x + lowbit(x);
   }
 }
 ll getsum1(int x) {  // a[1]……a[x]的和
   ll ans = ;
   while (x >= ) {
     ans = ans + c1[x];
     x = x - lowbit(x);
   }
   return ans;
 }
 ll getsum2(int x) {  // a[1]……a[x]的和
   ll ans = ;
   while (x >= ) {
     ans = (ans + c2[x]);
     x = x - lowbit(x);
   }
   return ans;
 }
 ll getsum3(int x) {  // a[1]……a[x]的和
   ll ans = ;
   while (x >= ) {
     ans = (ans + c3[x]);
     x = x - lowbit(x);
   }
   return ans;
 }
 int main()
 {
     scanf("%d",&n);
     string s;
     cin>>s;
     ll pos = ,ans = ,cnt = ;
     for(int i = ;i<n;i++){
         if(s[i] == '') add1(i+,),add2(i+,i+);
     }
     ll t = ;
     for(int i = ;i<n;i++){
         if(s[i] == '') cnt++,t = (t + i + ),ans = ( cnt*(i+) - t ),add3(i+,ans);
     }   //  pos = (pos + i)%mod,cnt++,ans = (ans + cnt*i-pos)%mod;
     int m;scanf("%d",&m);
     printf("%lld\n",getsum3(n)%mod);
     while(m--){
         ll q,pos;
         cin>>q>>pos;
         if(q == ){
             add1(pos,);
             add2(pos,pos);
             ll tmp = getsum1(pos)*pos-getsum2(pos);
             tmp = ( tmp + (getsum2(n)-getsum2(pos) - (getsum1(n)-getsum1(pos))*pos)) ;
             add3(pos,tmp);
         }
         else{
             ll tmp = getsum1(pos)*pos-getsum2(pos);
             tmp = ( tmp + (getsum2(n)-getsum2(pos) - (getsum1(n)-getsum1(pos))*pos)) ;
             add3(pos,-tmp);
             add1(pos,-);
             add2(pos,-(pos));
         }
         ll ans = getsum3(n)%mod;
         printf("%lld\n",ans);
     }
     return ;
 }