题面
已知一个数列,你需要进行下面两种操作:
1.将某区间每一个数加上x
2.求出某区间每一个数的和
题解
区间修改+区间查询
线段树模板
#include<iostream>
#define maxn 100010
#define lch p<<1
#define rch p<<1|1
using namespace std;
typedef long long LL;
struct node{
LL val, addmark;
}sgt[maxn<<2];
void pushdown(LL p, LL l, LL r){
if(sgt[p].addmark != 0){
LL t = sgt[p].addmark, m = l+r>>1;
sgt[lch].addmark += t;
sgt[rch].addmark += t;
sgt[lch].val += t*(m-l+1);
sgt[rch].val += t*(r-m);
sgt[p].addmark = 0;
}
}
void update(LL p, LL l, LL r, LL L, LL R, LL v){
if(l > R || r < L)return ; //越界返回
if(L <= l && R >= r){
sgt[p].addmark += v;
sgt[p].val += v*(r-l+1);
return ;
}
pushdown(p, l, r);//记得pushdown 的位置 每次递归前
LL m = l+r>>1;
update(lch, l, m, L, R, v);
update(rch, m+1, r, L, R, v);
sgt[p].val = sgt[lch].val + sgt[rch].val;
}
LL query(LL p, LL l, LL r, LL L, LL R){
if(l > R || r < L)return 0; //越界返回
if(L <= l && R >= r)return sgt[p].val;
pushdown(p,l,r);
LL m = l+r>>1, ans = 0;
ans += query(lch, l, m, L , R);
ans += query(rch, m+1, r, L , R);
return ans;
}
int main(){
LL n, m;
cin>>n>>m;
for(int i = 1; i <= n; i++){
LL x; cin>>x;
update(1, 1, n, i, i, x);
}
for(int i = 1; i <= m; i++){
LL op; cin>>op;
if(op == 1){
LL x, y, k; cin>>x>>y>>k;
update(1,1,n,x,y,k);
}else{
LL x, y; cin>>x>>y;
cout<<query(1, 1, n, x, y)<<"
";
}
}
return 0;
}