題目鏈接

loj

題解

感謝珂神的指導orz

觀察式子\(i \times j \equiv 1 \pmod m\)，顯然\(i,j\)是模\(m\)意義下成對的逆元，只需統計模\(m\)意義下存在逆元的數的個數，即與\(m\)互質的數的個數\(\varphi(m)\)
每對逆元的連邊有兩種情況，記逆元對數為\(x\)，則方案數為\(2^x\)

真的完了嗎？難點才剛開始
模\(m\)意義下有的數逆元為本身！此時不能計入答案
所以我們還需求模\(m\)意義下逆元為本身的數的個數

重新理解一下中國剩余定理，本質是環同構
令\(m = a_1a_2a_3\dots\)
其中\(a_i = p_i^{k_i}\)

環同樣滿足加減乘
所以我們需要求出在模每個\(a_i\)下平方等於\(1\)的數的個數，乘起來即可
\[ \begin{aligned} x^2 \equiv 1 \pmod m \(x + 1)(x - 1) \equiv 0 \pmod m \end{aligned} \]

所以我們只需\((x + 1)(x - 1)\)

#include<algorithm>
#include<iostream>
 

#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define res register
using namespace std;
const int maxn = 10000005,maxm = 100005,INF = 1000000000,P = 311021;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int p[maxn],pi,isn[maxn],phi[maxn],N = 10000000;
int f[maxn],bin[maxn],tw[maxn],tot[maxn],all[maxn];
void init(){
    phi[1] = 1;
    for (res int i = 2; i <= N; i++){
        if (!isn[i]){
            p[++pi] = i,phi[i] = i - 1;
            if (i == 2) tw[i] = 1;
            else tot[i] = 1;
        }
        for (res int j = 1; j <= pi && i * p[j] <= N; j++){
            isn[i * p[j]] = true;
            if (i % p[j] == 0){
                phi[i * p[j]] = phi[i] * p[j];
                if (p[j] == 2){
                    tw[i * p[j]] = tw[i] + 1;
                    tot[i * p[j]] = tot[i];
                }
                else{
                    tw[i * p[j]] = tw[i];
                    tot[i * p[j]] = tot[i];
                }
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
            if (p[j] == 2){
                tot[i * p[j]] = tot[i];
                tw[i * p[j]] = tw[i] + 1;
            }
            else{
                tot[i * p[j]] = tot[i] + 1;
                tw[i * p[j]] = tw[i];
            }
        }
    }
    bin[0] = 1;
    for (res int i = 1; i <= N; i++)
        bin[i] = (bin[i - 1] << 1),bin[i] >= P ? (bin[i] -= P) : 0;
    for (res int i = 1; i <= N; i++){
        all[i] = bin[tot[i]];
        if (tw[i] == 2) all[i] = all[i] * 2 % P;
        else if (tw[i] > 2) all[i] = all[i] * 4 % P;
    }
    for (res int i = 1; i <= N; i++){
        f[i] = f[i - 1] + bin[(phi[i] - all[i]) >> 1],f[i] >= P ? (f[i] -= P) : 0;
    }
}
int c[maxn],ci;
int main(){
    init();
    int q = read(),l,r;
    while (q--){
        l = read(); r = read();
        printf("%d\n",(f[r] - f[l - 1] + P) % P);
    }
    return 0;
}

loj Snakes 的 Naïve Graph 【數論】