题解 | 信息学奥赛一本通 樱花

思路

先考虑.
也就是.考虑..因为互质,所以.所以当且仅当且时满足条件.考虑把质因数分解,每个质因数分给和,或者和,或者全部给.形式化地,,.
改成也就是.复杂度.(反正复杂度都这样了还写线性筛干嘛qwq).

代码

#include<bits/stdc++.h>
using namespace std;
#define i64 long long
#define fp( i, b, e ) for ( int i(b), I(e); i <= I; ++i )
#define fd( i, b, e ) for ( int i(b), I(e); i >= I; --i )
#define go( i, b ) for ( int i(b), v(to[i]); i; v = to[i = nxt[i]] )
template<typename T> inline void cmax( T &x, T y ){ x < y ? x = y : x; }
template<typename T> inline void cmin( T &x, T y ){ y < x ? x = y : x; }
//#define getchar() ( p1 == p2 && ( p1 = bf, p2 = bf + fread( bf, 1, 1 << 21, stdin ), p1 == p2 ) ? EOF : *p1++ )
char bf[1 << 21], *p1(bf), *p2(bf);
template<typename T>
inline void read( T &x ){ char t(getchar()), flg(0); x = 0;
    for ( ; !isdigit(t); t = getchar() ) flg = t == '-';
    for ( ; isdigit(t); t = getchar() ) x = x * 10 + ( t & 15 );
    flg ? x = -x : x;
}

clock_t t_bg, t_ed;
int N, ans(1);
bool vis[1000005];
const int mod = 1e9 + 7;

signed main(){
    t_bg = clock();
    read(N);
    fp( i, 2, N ) if ( !vis[i] ){
        for ( int j = i + i; j <= N; j += i ) vis[j] = 1;
        int c(0); for ( i64 j = i; j <= N; j *= i ) c += N / j;
        ans = 1ll * ans * ( c << 1 | 1 ) % mod;
    } printf( "%d\n", ans );
    t_ed = clock();
    fprintf( stderr, "\n========info========\ntime : %.3f\n====================\n", (double)( t_ed - t_bg ) / CLOCKS_PER_SEC );
    return 0;
}