P1829 [国家集训队]Crash的数字表格 / JZPTAB

题意：

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

题解：

跟这个题P3911 最小公倍数之和很相近，但是本题数据范围大


tmp是整数分块过程中d在[l,r]这段区间的累加

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

代码：

#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();
}