题目链接
题解
做不动其他的毒瘤组合数学只能来写点水题了QAQQQQ
对于挡住i,j的点数显然是gcd(i,j)
那么就是求
(2 imes sum_i^n sum_j^m gcd(i,j) -n imes m)
枚举带约数(p)
令
[ans=sum_{p=1}^{n}p*sum_{i=1}^{n}sum_{j=1}^{m}[gcd(i,j)=p]
]
$$ans=sum_{p=1}^{n}p*sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}[gcd(i,j)=1]$$
[ans=sum_{p=1}^{n}p*sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}sum_{d|i}sum_{d|j}mu(d)
]
枚举约数d
[ans=sum_{p=1}^{n}p*sum_{d=1}^{lfloorfrac{n}{p}
floor}mu(d)lfloorfrac{n}{pd}
floorlfloorfrac{m}{pd}
floor
]
答案就是(2 imes ans - n imes m)
复杂度(O(nsqrt{n}))
代码
#include<cstdio>
#include<algorithm>
inline int read() {
int x = 0,f = 1;
char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar(); }
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
return x * f;
}
#define int long long
const int maxn = 100007;
int n,m,mu[maxn],prime[maxn],num;bool p[maxn];
void get_mu() {
mu[1] = 1; int k = maxn - 7;
for(int i = 2;i <= k;++ i) {
if(!p[i]) mu[i] = -1,prime[++ num] = i;
for(int j = 1;j <= num && prime[j] * i <= k;++ j) {
p[prime[j] * i] = 1;
if(i % prime[j] == 0) break;
mu[prime[j] * i] = mu[i] * -1;
}
mu[i] += mu[i - 1];
}
}
inline int calc(int x) {
int a = n / x ,b = m / x,ret = 0;
for(int i = 1,last;i <= a;i = last + 1) {
last = std::min(a / (a / i),b / (b / i));
ret += (mu[last] - mu[i - 1]) * (a / i) * (b / i);
}
return ret;
}
main() {
get_mu();
n = read(),m = read();
if(n > m) std::swap(n,m);
int ans = 0;
for(int i = 1;i <= n;++ i) ans += 2 * i * calc(i);
printf("%lld
",ans - n * m);
return 0;
}