题目
弱化版题目的传送门（【BZOJ2154】Crash的数字表格）
思路&解法
题目是要求： (sumlimits_{i = 1}^{n}sumlimits_{j = 1}^{m}lcm(i, j))
于是我们可以把式子化成这样：
[sum_{i = 1}^{n}sum_{j = 1}^{m}frac{ij}{gcd(i, j)}
]
然后我们枚举gcd
[sum_{i = 1}^{n}sum_{j = 1}^{m} sum_{k = 1}^{min(n, m)}frac{ij}{k}[gcd(i, j) == k]
]
我们再把式子换一下
[sumlimits_{k = 1}^{min(n, m)}{frac{1}{k}sumlimits_{i = 1}^{n}sumlimits_{j = 1}^{m}ij[gcd(i, j) == k]}
]
[sumlimits_{k = 1}^{min(n, m)}{frac{1}{k}sumlimits_{i = 1}^{lfloor{frac{n}{k}}
floor}sumlimits_{j = 1}^{lfloorfrac{m}{k}
floor}ijk^2[gcd(i, j) == 1]}
]
[sum^{min(n, m)}_{k = 1}ksum_{i= 1}^{lfloor{frac{n}{k}}
floor}sum_{j = 1}^{lfloor{frac{m}{k}}
floor}ij[gcd(i, j) ==1]
]
反演一下
[sum_{k}^{min(n, m)} k sum_{d = 1}^{min(lfloor {frac{n}{k}}
floor, lfloor {frac{m}{k}}
floor)}mu(d) imes sum_{i|d}sum_{j|d} ij
]
[sum_{k}^{min(n, m)} ksum_{d = 1}^{min(lfloor {frac{n}{k}}
floor, lfloor {frac{m}{k}}
floor)} mu(d) imes d^2sum_{i=1}^{lfloor{frac{n}{kd}}
floor}sum_{j=1}^{lfloorfrac{m}{kd}
floor} ij
]
[sum_{k}^{min(n, m)} ksum_{d = 1}^{min(lfloor {frac{n}{k}}
floor, lfloor {frac{m}{k}}
floor)} mu(d)d^2 F(lfloor{frac{n}{kd}}
floor, lfloor{frac{m}{kd}}
floor)
]
其中$$F(n, m) = {nm(n+1)(m+1)over 4}$$
继续优化
[sum_{T=1}^{min(n, m)}{F(lfloor{frac{n}{T}}
floor, lfloor{frac{m}{T}}
floor)}sum_{d|T}mu(d)d^2{frac{T}{d}}
]
后面的(sumlimits_{d|T}mu(d)d^2{frac{T}{d}})的前缀和很容易求
代码
【BZOJ2693】jzptab
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 100000009LL;
const int N = 10000010;
int p[N], total;
bool vis[N];
LL g[N];
void init()
{ g[1] = 1;
for (int i = 2; i <= 10000000; i++)
{ if (!vis[i]) p[++total] = i, g[i] = (LL) (1 - i + mod) % mod;
for (int j = 1; j <= total && i * (LL) p[j] <= 10000000; j++)
{ vis[i * p[j]] = 1;
if (i % p[j] == 0) { g[i * p[j]] = g[i]; break; }
else g[i * p[j]] = (g[i] * g[p[j]]) % mod;
}
}
for (int i = 2; i <= 10000000; i++) g[i] = (g[i] * i + g[i-1]) % mod;
}
LL Get(int n) { return ((LL) n * (LL) (n+1) / 2LL) % mod; }
LL Sum(int n, int m) { return (Get(n) * Get(m)) % mod; }
LL Calc(int n, int m)
{ int last = 0;
LL Ans = 0;
for (int i = 1; i <= min(n, m); i = last+1)
{ last = min(n / (n/i), m / (m/i));
Ans = (Ans + (Sum(n/i, m/i) * (g[last] - g[i-1])) % mod) % mod;
}
return (Ans + mod) % mod;
}
int main()
{ init();
int T;
scanf("%d", &T);
while (T--)
{ int n, m;
scanf("%d %d", &n, &m);
printf("%lld
", Calc(n, m));
}
return 0;
}
【BZOJ2154】Crash的数字表格
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 20101009LL;
const int N = 10000010;
int p[N], total;
bool vis[N];
LL g[N];
void init()
{ g[1] = 1;
for (int i = 2; i <= 10000000; i++)
{ if (!vis[i]) p[++total] = i, g[i] = (LL) (1 - i + mod) % mod;
for (int j = 1; j <= total && i * (LL) p[j] <= 10000000; j++)
{ vis[i * p[j]] = 1;
if (i % p[j] == 0) { g[i * p[j]] = g[i]; break; }
else g[i * p[j]] = (g[i] * g[p[j]]) % mod;
}
}
for (int i = 2; i <= 10000000; i++) g[i] = (g[i] * i + g[i-1]) % mod;
}
LL Get(int n) { return ((LL) n * (LL) (n+1) / 2LL) % mod; }
LL Sum(int n, int m) { return (Get(n) * Get(m)) % mod; }
LL Calc(int n, int m)
{ int last = 0;
LL Ans = 0;
for (int i = 1; i <= min(n, m); i = last+1)
{ last = min(n / (n/i), m / (m/i));
Ans = (Ans + (Sum(n/i, m/i) * (g[last] - g[i-1])) % mod) % mod;
}
return (Ans + mod) % mod;
}
int main()
{ init();
int n, m;
scanf("%d %d", &n, &m);
printf("%lld
", Calc(n, m));
return 0;
}
一些其他的东西
弱化版题目可以(O(n))过， 然而我是用(O(sqrt{n}))的算法做的， 而且达到了惊人的18s， 比(O(n))的解法慢多了。我哪里写锉了。。。。。。