(color{#0066ff}{ 题目描述 })
给定整数N,求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对.
(color{#0066ff}{输入格式})
一个整数N
(color{#0066ff}{输出格式})
答案
(color{#0066ff}{输入样例})
4
(color{#0066ff}{输出样例})
4
(color{#0066ff}{数据范围与提示})
对于样例(2,2),(2,4),(3,3),(4,2)
1<=N<=10^7
(color{#0066ff}{ 题解 })
[sum_{pin prime} sum_{i=1}^n sum_{j=1}^n [gcd(i,j)==p]
]
[sum_{pin prime} sum_{i=1}^{lfloorfrac n p
floor} sum_{j=1}^{lfloorfrac n p
floor} [gcd(i,j)==1]
]
拿(varphi) xjb统计一下就行了
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 1e7 + 10;
int pri[maxn], tot, n;
LL phi[maxn];
bool vis[maxn];
signed main() {
n = in();
phi[1] = 1;
for(int i = 2; i <= n; i++) {
if(!vis[i]) pri[++tot] = i, phi[i] = i - 1;
for(int j = 1; j <= tot && (LL)i * pri[j] <= n; j++) {
vis[i * pri[j]] = true;
if(i % pri[j] == 0) {
phi[i * pri[j]] = phi[i] * pri[j];
break;
}
else phi[i * pri[j]] = phi[i] * (pri[j] - 1);
}
}
for(int i = 2; i <= n; i++) phi[i] += phi[i - 1];
LL ans = 0;
for(int i = tot; i >= 1; i--) {
ans += (phi[n / pri[i]] << 1LL) - 1;
}
printf("%lld
", ans);
return 0;
}