P1829 [國家集訓隊]Crash的數字表格 / JZPTAB

題意：

求${\sum }_{i=1}^n{\sum }_{j=1}^{m}lcm(i,j)$
1<=n<=m<=1e7
結果mod20101009

題解：

跟這個題P3911 最小公倍數之和很相近，但是本題數據范圍大


tmp是整數分塊過程中d在[l,r]這段區間的累加

ll tmp= ((1ll * r * (r + 1) / 2) - (1ll * (l - 1) * l / 2)) % mod;

代碼：

#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; void read(){}; template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar) {x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...); } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin); #endif } void Time_test() { #ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn= 2e7 + 9; const int mod= 20101009; int prime[maxn], mu[maxn]; int vis[maxn]; ll sum[maxn]; int cnt= 0; void get_mu(int N) {mu[1]= 1;vis[1]= vis[0]= 1;for (int i= 2; i <= N; i++) {if (!vis[i]) {prime[++cnt]= i;mu[i]= -1;}for (int j= 1; j <= cnt && i * prime[j] <= N; j++) {vis[i * prime[j]]= 1;if (i % prime[j] == 0)break;mu[i * prime[j]]= -mu[i];}}for (int i= 1; i <= N; i++) {sum[i]= (sum[i - 1] + 1ll * i * i % mod * mu[i] % mod) % mod;} } ll f(int x, int y) {ll ans= (1ll * x * (x + 1) / 2) % mod * (1ll * y * (y + 1) / 2 % mod) % mod;return ans % mod; } ll Sum(int x, int y) {ll ans= 0;for (int l= 1, r; l <= min(x, y); l= r + 1) {r= min(x / (x / l), y / (y / l));ans= (ans + 1ll * (sum[r] - sum[l - 1] + mod) % mod * f(x / l, y / l) % mod) % mod;}//cout << "ans=" << ans << endl;return ans % mod; } ll poww(ll a, ll b) {ll ans= 1;while (b) {if (b & 1)ans= ans * a % mod;a= a * a % mod;b>>= 1;}return ans % mod; } int main() {get_mu(10000002);//rd_test();int n, m;read(n, m);int minn= min(n, m);ll ans= 0;for (int l= 1, r; l <= minn; l= r + 1) {r= min(n / (n / l), m / (m / l));ll tmp= ((1ll * r * (r + 1) / 2) - (1ll * (l - 1) * l / 2)) % mod;ans= (ans + tmp * Sum(n / l, m / l) % mod) % mod;//cout << ans % mod << endl;}cout << ans % mod;//Time_test(); }

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