Description
我们知道,从区间[L,H](L和
Solution
2.1 递推
先把左右区间同除k,然后考虑求互质的数列。
用
然后从后往前递推并减去其倍数的贡献即可。
至于为什么必须是选出的数不全相同,这是因为有可能选出的数全部相同、但是公约数并不在范围之内。
另外,如果k在左右区间内,注意将答案加上
2.2 莫比乌斯反演
待填坑。。。
Code
3.1
/**************************
* Au: Hany01
* Date: Jan 18th, 2018
* Prob: bzoj3930 & cqoi2015 递推
* Email: hany01@foxmail.com
**************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
inline int read()
{
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
inline void File()
{
#ifdef hany01
freopen("bzoj3930.in", "r", stdin);
freopen("bzoj3930.out", "w", stdout);
#endif
}
const int maxn = 100005;
int n, k, l, r, len, f[maxn], tmp;
inline int Pow(LL a, int b) {
LL Ans = 1;
for ( ; b; a = a * a % Mod, b >>= 1) if (b & 1) (Ans *= a) %= Mod;
return Ans;
}
inline void mod(int &x) { if (x < 0) x += Mod; }
inline int calc(int x) { return floor(r * 1.0 / x) - floor((l - 1) * 1.0 / x); }
int main()
{
File();
n = read(), k = read(), l = read(), r = read();
l = ceil(l * 1.0 / k), r = floor(r * 1.0 / k), len = r - l;
Fordown(i, len, 1) {
tmp = calc(i),
f[i] = Pow(tmp, n) - tmp;
for (register int j = 2; j * i <= len; ++ j) mod(f[i] -= f[j * i]);
}
if (l <= 1) ++ f[1];
if (f[1] == Mod) f[1] = 0;
printf("%d\n", f[1]);
return 0;
}
//雪纷纷,掩重门,不由人不断魂,瘦损江梅韵。
// -- 关汉卿《大德歌·冬》