- 题目链接:http://codeforces.com/contest/920/problem/F
- 题意:定义一个函数D(x) = x 的因数数量总和。给你n个数 a[1] ~ a[n]。给你两种命令
- 将下标在[l, r] 区间内的数x,全部变为D(x)
- 查询[l,r]的区间和
- 思路:容易得知,D(1) = 1, D(2)=2, 所以,以最底层数为更新对象,当某个区间内的最底层数全为1或者2时,可以直接跳过。
- 不足:寻找某个数的因数个数时方法太暴力,据说可以用Eratosthenes sieve找,博主正在尝试
#include <bits/stdc++.h>
#define pi acos(-1)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL ll_INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 3e5 + 10;
const int maxm = 1e6 + 10;
const int mod = 1e9 + 7;
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
LL sum[maxn<<2];
int a[maxn], fla[maxm];
bool skip[maxn<<2];
int n, m;
void PushUp(int rt)
{
sum[rt] = sum[rt<<1]+sum[rt<<1|1];
skip[rt] = skip[rt<<1] && skip[rt<<1|1];
}
void Build(int l, int r, int rt)
{
if(l==r){
sum[rt]=a[l];
if(sum[rt]<=2) skip[rt] = true;
return;
}
int mid = (l+r)>>1;
Build(l, mid, rt<<1);
Build(mid+1, r, rt<<1|1);
PushUp(rt);
}
void Update(int L, int R, int l, int r, int rt)
{
if(skip[rt]) return ;
if(l==r){
sum[rt] = fla[sum[rt]];
if(sum[rt]<=2) skip[rt] = true;
return ;
}
int mid = (l+r)>>1;
if(L <= mid) Update(L, R, l, mid, rt<<1);
if(R > mid) Update(L, R, mid+1, r, rt<<1|1);
PushUp(rt);
}
LL Query(int L, int R, int l, int r, int rt)
{
if(L<=l && r<=R){
return sum[rt];
}
int mid = (l+r)>>1;
LL ans=0;
if(L<=mid) ans += Query(L, R, l, mid, rt<<1);
if(R> mid) ans += Query(L, R, mid+1, r, rt<<1|1);
return ans;
}
void init()
{
memset(skip, false, sizeof(skip));
memset(fla, 0, sizeof(fla));
for(int i=1; i<=maxm; i++){
for(int j=i; j<=maxm; j+=i)
fla[j]++;
}
}
int main()
{
init();
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) scanf("%d", &a[i]);
Build(1, n, 1);
int t, l, r;
while(m--){
scanf("%d%d%d", &t, &l, &r);
if(t==1) Update(l, r, 1, n, 1);
else printf("%I64d\n", Query(l, r, 1, n, 1));
}
}