Another Array of Orz Pandas

Another Array of Orz Pandas

题目链接：http://acm.xidian.edu.cn/problem.php?id=1187

线段树

线段树维护区间和以及区间内各个数平方和，对于每一个询问ans=(sum²-pow_sum)/2

代码如下：

 #include<cstdio>
 #include<cstring>
 #define lson (x<<1)
 #define rson (x<<1|1)
 #define mid ((l+r)>>1)
 #define N 100007
 typedef long long LL;
 struct nod{
     LL sum,lazy,or2zds;
 }a[N<<];
 const LL mod=1e9+;
 const LL Max=1e9;
 LL n,m;
 void push_up(LL x){
     a[x].sum=(a[lson].sum+a[rson].sum)%mod;
     a[x].or2zds=(a[lson].or2zds+a[rson].or2zds)%mod;
 }
 void push_down(LL x,LL l,LL r){
     a[lson].or2zds=(a[lson].or2zds+(a[x].lazy*a[x].lazy)%mod*(mid+-l)+*a[x].lazy*a[lson].sum+mod)%mod;
     a[lson].sum=(a[lson].sum+a[x].lazy*(mid+-l))%mod;
     a[lson].lazy=(a[lson].lazy+a[x].lazy)%mod;
     a[rson].or2zds=(a[rson].or2zds+(a[x].lazy*a[x].lazy)%mod*(r-mid)+*a[x].lazy*a[rson].sum+mod)%mod;
     a[rson].sum=(a[rson].sum+a[x].lazy*(r-mid))%mod;
     a[rson].lazy=(a[rson].lazy+a[x].lazy)%mod;
     a[x].lazy=;
 }
 void add(LL x,LL l,LL r,LL cl,LL cr,LL v){
     if(cl<=l&&r<=cr){
         a[x].or2zds=(a[x].or2zds+(v*v)%mod*(r-l+)%mod+*v*a[x].sum)%mod;
         a[x].sum=(a[x].sum+v*(r-l+))%mod;
         a[x].lazy=(a[x].lazy+v)%mod;
         return;
     }
     if(a[x].lazy!=)push_down(x,l,r);
     if(cl<=mid)add(lson,l,mid,cl,cr,v);
     if(mid<cr)add(rson,mid+,r,cl,cr,v);
     push_up(x);
 }
 void query(LL x,LL l,LL r,LL ql,LL qr,LL &sum,LL &sum2){
     if(ql<=l&&r<=qr){
         sum=(sum+a[x].sum)%mod;
         sum2=(sum2+a[x].or2zds)%mod;
         return;
     }
     if(a[x].lazy!=)push_down(x,l,r);
     if(ql<=mid)query(lson,l,mid,ql,qr,sum,sum2);
     if(mid<qr)query(rson,mid+,r,ql,qr,sum,sum2);
 }
 int main(void){
     while(~scanf("%lld%lld",&n,&m)){
         memset(a,,sizeof(a));
         LL l,r,op,k;
         for(LL i=;i<m;i++){
             scanf("%lld",&op);
             if(op==){
                 scanf("%lld%lld%lld",&l,&r,&k);
                 add(,,n,l,r,k);
             }else {
                 scanf("%lld%lld",&l,&r);
                 LL sum=,sum2=;
                 query(,,n,l,r,sum,sum2);
                 printf("%lld\n",(sum*sum%mod-sum2+mod)%mod*%mod);
             }
         }
     }
     return ;
 }