• P2260 [清华集训2012]模积和(数论分块)


      1 #include <iostream>
      2 #include <cstdio>
      3 #include <algorithm>
      4 #include <vector>
      5 #include<set>
      6 #include <map>
      7 #include <string>
      8 using namespace std;
      9 
     10 #define ll long long
     11 #define pb push_back
     12 
     13 const ll MOD = 19940417;
     14 const int N = 1e6;
     15 
     16 
     17 ll inv2, inv6;
     18 
     19 // ll gcd( ll a, ll b)
     20 // {
     21 //     return b == 0 ? a : gcd(b, a % b);
     22 // }
     23 
     24 ll ksm(ll a, ll b, ll mod)
     25 {
     26     ll res = 1;
     27     while(b){
     28         if(b & 1) res = (res * a) % mod;
     29         a = a * a % mod;
     30         b >>= 1;
     31     }
     32     return res;
     33 }
     34 
     35 inline ll fx2(ll n)
     36 {
     37     return n * (n + 1) % MOD * inv2 % MOD;
     38 }
     39 
     40 ll fun1(ll n)
     41 {
     42     ll l, r, sum1, sum2;
     43     sum1 = sum2 = 0;
     44     sum1 = n * n % MOD;
     45     for(l = 1, r = 0; l <= n; l = r + 1){
     46         r = n / (n / l);
     47         sum2 = (sum2 + (fx2(r) - fx2(l - 1) + MOD) * (n / l) % MOD) % MOD;
     48     }
     49     return (sum1 - sum2 + MOD) % MOD;
     50 }
     51 
     52 ll fun2(ll n, ll m)
     53 {
     54     ll l, r, sum1, sum2;
     55     sum1 = sum2 = 0;
     56     //cout << n << " " << m << endl;
     57     for(l = 1, r = 0; l <= n; l = r + 1){
     58         r = n / (n / l);
     59         sum1 = (sum1 + (fx2(r) - fx2(l - 1) + MOD) * (n / l)) % MOD;
     60     }
     61     sum1 = m % MOD * sum1 % MOD;
     62 
     63     for(l = 1, r = 0; l <= n; l = r + 1){
     64         r = min(n, m / (m / l));
     65         sum2 = (sum2 + (fx2(r) - fx2(l - 1) + MOD) * (m / l)) % MOD;
     66     }
     67     sum2 = n % MOD * sum2 % MOD;
     68 
     69     return (sum1 + sum2) % MOD;
     70 }
     71 
     72 inline ll fx6(ll n)
     73 {
     74     return n * (n + 1) % MOD * (2 * n + 1) % MOD * inv6 % MOD;
     75 }
     76 
     77 ll fun3(ll n, ll m)
     78 {   
     79     ll l, r, sum;
     80     sum = 0;
     81     for(l = 1, r = 0; l <= n; l = r + 1){
     82         //取右边小的使得值正确,如果取到了右边大的,会导致a或者b变小
     83         r = min(n / (n / l), m / (m / l));
     84         // /r = n / (n / l);
     85         ll a = n / l;
     86         ll b = m / l;
     87         sum = (sum + (fx6(r) - fx6(l - 1) + MOD) * a % MOD * b % MOD) % MOD;     
     88     }
     89     return sum % MOD;
     90 }
     91 
     92 
     93 
     94 ll exgcd(ll a, ll b, ll &x, ll &y)
     95 {
     96 
     97     if(b == 0){//推理1,终止条件
     98         x = 1;
     99         y = 0;
    100         return a;
    101     }
    102     ll r = exgcd(b, a%b, x, y);
    103     //先得到更底层的x2,y2,再根据计算好的x2,y2计算x1,y1。
    104     //推理2,递推关系
    105     ll t = y;
    106     y = x - (a/b) * y;
    107     x = t;
    108     return r;
    109 }
    110 
    111 void inv()
    112 {
    113 
    114     //ax同余1(mod prime)  求最小正整数解
    115     // ax + by = n ---> ax' + by' = gcd(a, b)
    116     // k = gcd(a, b)  k*(ax' + by') = n
    117     //以下 a, b, x, y对应上面 a, b, x', y'
    118     ll x, y;
    119     //以下求出mod不为质数的逆元
    120     // a / b % mod c
    121     // a / b * b^-1 ≡ x * b^-1(mod c)
    122     // b * b^-1 ≡ 1 (mod c)
    123     // b * b^-1 + y * c  = 1
    124     // a * x + b * y = 1
    125     exgcd(2, MOD, x, y);
    126     inv2 = (x + MOD) % MOD;
    127     exgcd(6, MOD, x, y);
    128     inv6 = (x + MOD) % MOD;
    129 
    130     // cout << "inv2 = " << (inv2 + MOD) % MOD << endl;
    131     // cout << "inv6 = " << (inv6 + MOD) % MOD << endl;
    132 }
    133 
    134 void solve()
    135 {   
    136 
    137     //cout << gcd(2, MOD) << " " << gcd(6, MOD) << endl;
    138     //cout << inv2 << " " << inv6 << endl;
    139     inv();
    140    // cout << inv2 << " " << inv6 << endl;
    141     ll n, m;
    142     cin >> n >> m;
    143     if(n > m) swap(n, m);
    144     ll ans1 = (fun1(n) * fun1(m)) % MOD;
    145     ll ans2 = (n * n % MOD * m % MOD - fun2(n, m) + fun3(n, m) + MOD) % MOD;
    146     ll ans = (ans1 - ans2 + MOD) % MOD;
    147 
    148     cout << ans << endl;
    149 }
    150 
    151 int main()
    152 {   
    153     ios::sync_with_stdio(false);
    154     cin.tie(0);
    155     cout.tie(0);
    156     solve();
    157     //int ok = 0;
    158     //cout << "ok" << endl;
    159     return 0;
    160 }
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  • 原文地址:https://www.cnblogs.com/SSummerZzz/p/13445503.html
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