该题属于经典题,用到了延迟标记,意思是当你需要时才去更新,而不是一次更新到底,这样的更新和查询都是logn.
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn = 110000;
LL a[maxn << 2]={0};
LL st[maxn << 2];
void GetSum(int rt){ st[rt] = st[rt << 1] + st[rt << 1 | 1]; }
void MoveDown(int rt, int len){
if(a[rt]){
a[rt << 1] += a[rt];
a[rt << 1 | 1] += a[rt];
st[rt << 1] += a[rt] * (len - (len >> 1 ));
st[rt << 1 | 1] += a[rt] * (len >> 1);
a[rt] = 0;
}
}
void BuildTree( int rt, int l, int r){
if(l == r){
scanf("%I64d",&st[rt]);
return ;
}
int mid = (l + r) >> 1;
BuildTree( rt << 1, l, mid);
BuildTree( rt << 1 | 1, mid + 1, r);
GetSum(rt);
}
void Update(int L, int R, int w, int rt, int l, int r){
if(L <= l && R >= r){
st[rt] += w * (r - l + 1);
a[rt] += w;
return ;
}
MoveDown( rt, r - l + 1);
int mid = (l + r) >> 1;
if(L <= mid) Update( L, R, w, rt << 1, l, mid);
if(R > mid) Update( L, R, w, rt << 1 | 1, mid + 1, r);
GetSum(rt);
}
LL Query(int L, int R, int rt, int l, int r){
if(L <= l && R >= r)
return st[rt];
LL ans = 0;
MoveDown(rt , r - l + 1);
int mid = (r + l ) >> 1;
if(L <= mid ) ans += Query( L, R, rt << 1, l, mid);
if(R > mid ) ans += Query( L, R, rt << 1 | 1, mid + 1, r);
return ans;
}
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m)){
BuildTree(1, 1, n);
char s[2];
int x,y,z;
for(int i = 1; i <= m; ++i){
scanf("%s",s);
if(s[0] == 'C'){
scanf("%d %d %d",&x,&y,&z);
Update( x, y, z, 1, 1, n);
}
else {
scanf("%d %d",&x,&y);
printf("%I64d\n",Query( x, y, 1, 1, n));
}
}
}
return 0;
}