Time Limit: 15 Sec Memory Limit: 256 MB
Submit: 434 Solved: 262
[Submit][Status][Discuss]
Description
给出一个数字N
Input
第一行为一个正整数T，表示数据组数。
接下来T行为询问，每行包含一个正整数N。
T<=5000,N<=10^7
Output
按读入顺序输出答案。
Sample Input
1
10
Sample Output
136
HINT
Source
By FancyCoder
题解
[sum_{i=1}^n sum_{j=1}^n phi(gcd(i,j))
]
[=sum_{i=1}^n sum_{j=1}^n sum_{d=1}^n phi(d)[gcd(i,j) =d]
]
[=sum_{d=1}^nphi(d)sum_{i=1}^n sum_{j=1}^n[gcd(i,j)=d]
]
[=sum_{d=1}^nphi(d)sum_{i=1}^{lfloor frac{n}{d}
floor} sum_{j=1}^{lfloor frac{n}{d}
floor}[gcd(i,j)=1]
]
不难发现(sum_{i=1}^{lfloor frac{n}{d} floor} sum_{j=1}^{lfloor frac{n}{d} floor}[gcd(i,j)=1])其实就是两倍欧拉函数前缀和减去1（因为(1,1)和(1,1)重复）
于是线性筛(phi)，求前缀和即可
好像(n,n)改成(n,m)也可做，不过线筛起来挺（我）麻（不）烦（会）
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline long long max(long long a, long long b){return a > b ? a : b;}
inline long long min(long long a, long long b){return a < b ? a : b;}
inline void swap(long long &x, long long &y){long long tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9') c = ch, ch = getchar();
while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f;
const long long MAXN = 10000000;
long long p[MAXN + 1], bp[MAXN + 1], tot, phi[MAXN + 1];
void make_phi()
{
phi[1] = 1;
for(long long i = 2;i <= MAXN;++ i)
{
if(!bp[i]) phi[i] = i - 1, p[++ tot] = i;
for(long long j = 1;j <= tot && i * p[j] <= MAXN;++ j)
{
bp[i * p[j]] = 1;
if(i % p[j] == 0)
{
phi[i * p[j]] = phi[i] * p[j];
break;
}
phi[i * p[j]] = phi[i] * (p[j] - 1);
}
}
for(long long i = 1;i <= MAXN;++ i) phi[i] += phi[i - 1];
}
long long t, n;
int main()
{
make_phi();
read(t);
for(;t;--t)
{
read(n);
long long ans = 0, r;
for(long long d = 1;d <= n;++ d)
{
r = min(n/(n/d), n);
ans += (phi[r] - phi[d - 1]) * ((phi[n/d] << 1) - 1);
d = r;
}
printf("%lld
", ans);
}
return 0;
}