bzoj2752: [HAOI2012]高速公路(road)

题目链接

bzoj2752: [HAOI2012]高速公路(road)

题解

首先将每次修改和询问的r减1，把线段权值转化为点权值。

然后使用总和/总次数的方式计算期望。

第i个点(l≤i≤r)，它被选中的次数为\((i−l+1)∗(r−i+1)\)所以所求即为



线段树维护\(v[i]∗i∗i、v[i]∗i、v[i]\)的区间和即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define int long long
const int  maxn = 200007;
inline LL read() {
LL x = 0,f = 1;
char c = getchar();
while(c < '0' || c > '9'){if(c == '-')f = -1;  c = getchar();}
while(c <= '9' && c >= '0')x = x * 10 + c - '0',c = getchar();
return x * f;
}
#define ls rt << 1
#define rs rt << 1 | 1

struct Segtree{
int l,r;
LL sum[5],tag;
Segtree (){
sum[0] = sum[1] = sum[2] = 0;
tag = 0;
}
} t[maxn << 2];
LL pre[maxn][3],ans[3];
inline LL Pre (Segtree a,int i) {
return pre[a.r][i] - pre[a.l - 1][i];
}
inline void updata(int rt) {
for(int i = 0;i < 3;++ i) t[rt].sum[i] = t[ls].sum[i] + t[rs].sum[i];
}
inline void pushdown(int rt) {
if(t[rt].tag == 0) return;
t[ls].tag += t[rt].tag;
t[rs].tag += t[rt].tag;
for(int i=0;i<3;++i) {
t[ls].sum[i] += t[rt].tag * Pre(t[ls],i);
t[rs].sum[i] += t[rt].tag * Pre(t[rs],i);
}
t[rt].tag = 0;
}
void build(int rt,int l,int r) {
t[rt].l = l,t[rt].r = r;
if(l == r) return;
int mid = l + r >> 1 ;
build(ls,l,mid);
build(rs,mid+1,r);
return;
}
void modify(int rt,int tl,int tr,LL x) {
int l = t[rt].l,r = t[rt].r;
pushdown(rt);
if(l >= tl && tr >= r) {
t[rt].tag += x;
for(int i = 0;i < 3;++ i) t[rt].sum[i] += x * Pre(t[rt],i);
return;
}
int mid = l + r >> 1;
if(tl <= mid) modify(ls,tl,tr,x);
if(tr >  mid) modify(rs,tl,tr,x);
updata(rt);
return;
}
LL query(int rt,int tl,int tr,int i) {
int l = t[rt].l,r = t[rt].r;
pushdown(rt);
if(l >= tl && r <= tr) return t[rt].sum[i];
int mid = l + r >> 1;LL ret = 0;
if(tl <= mid) ret += query(ls,tl,tr,i);
if(tr >  mid) ret += query(rs,tl,tr,i);
return ret;
}
LL gcd(LL a,LL b) {
return b == 0ll ? a : gcd(b,a % b);
}
main() {
for(int i = 1;i <= 100000;++ i) {
pre[i][0] = pre[i-1][0] + 1ll;
pre[i][1] = pre[i - 1][1] + (LL) i;
pre[i][2] = pre[i - 1][2] + (LL) i * i;
}
int n = read(),m = read();
build(1,1,n-1);
char ch[10];
while(m --) {
scanf("%s",ch + 1);
int  l = read(),r = read();
if(ch[1] == 'C') {
LL x = read();
modify(1,l,r-1,x);
} else {
for(int i = 0;i < 3;++ i) ans[i] = query(1,l,r - 1,i);
LL a = (r - l * r) * ans[0] + (l + r - 1) * ans[1] - ans[2];
LL b = ((1ll * (r - l + 1) * (r - l)) >> 1),c = gcd(a,b);
cout<<a / c<<"/"<<b / c<<endl;
}
}
return 0;
}