传送门
题目描述
分析
假设
f
(
i
)
=
x
f(i) = x
f(i)=x,那么
i
m
o
d
l
c
m
(
1
,
2
,
3...
i
−
1
)
=
0
i \ mod\ lcm(1,2,3...i - 1) = 0
i mod lcm(1,2,3...i−1)=0 且
i
m
o
d
l
c
m
(
1
,
2
,
3...
i
)
!
=
0
i \ mod\ lcm(1,2,3...i) != 0
i mod lcm(1,2,3...i)!=0
处理一下
l
c
m
lcm
lcm,根据容斥原理计算一下个数即可
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
ll gcd(ll a, ll b) {return (b > 0) ? gcd(b, a % b) : a;}
ll f[N];
int main() {
int T;
read(T);
f[1] = 1;
while(T--){
ll x;
read(x);
ll res = 0;
for(int i = 2;;i++){
f[i] = f[i - 1] / gcd(f[i - 1], i) * i;
res = (res + i * (x / f[i - 1] - x / f[i]) % mod) % mod;
if(f[i] > x) break;
}
dl(res);
}
return 0;
}
/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/