杭电多校第十场 hdu6434 Count 欧拉函数打表 快速打表模板

Problem I. Count

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 42    Accepted Submission(s): 16

Problem Description

Multiple query, for each n, you need to get
n i-1
∑ ∑ [gcd(i + j, i - j) = 1]
i=1 j=1

 

Input

On the first line, there is a positive integer T, which describe the number of queries. Next there are T lines, each line give a positive integer n, as mentioned above.
T<=1e5, n<=2e7

 

Output

Your output should include T lines, for each line, output the answer for the corre- sponding n.

 

Sample Input

4
978
438
233
666

 

Sample Output

194041
38951
11065
89963

 

Source

2018 Multi-University Training Contest 10

 

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题意：求下面这个式子的值

n i-1
∑ ∑ [gcd(i + j, i - j) = 1]
i=1 j=1

分析：按照上面那个式子直接暴力打表容易得到上面式子的含义是1-n以内偶数的欧拉函数值加上奇数的欧拉函数值除以2的和（注意要快速打表，运用素数快速打表）

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 2*1e7+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll num[maxn], sum[maxn], prim[maxn];
int main() {
    ios::sync_with_stdio(0);
    ll T, n;
    memset(num, 0, sizeof num);
    num[1] = 1;
    ll id = 0;
    for( ll i = 2; i <= maxn-10; i ++ ) {
        if(!num[i]) {
            num[i] = i - 1; prim[id++] = i;
        }
        for( ll j = 0; j < id && prim[j]*i <= maxn-10; j ++ ) {
            if(i % prim[j]) {
                num[i*prim[j]] = num[i] * (prim[j]-1);
            }
            else {
                num[i*prim[j]] = num[i] * prim[j];
                break;
            }
        }
    }
    num[1] = 0;
    sum[1] = 0;
    for( ll i = 2; i <= maxn-10; i ++ ) {
        if( i%2 == 0 ) {
            sum[i] = sum[i-1] + num[i];
        } else {
            sum[i] = sum[i-1] + num[i]/2;
        }
    }
    scanf("%lld",&T);
    while( T -- ) {
        scanf("%lld",&n);
        printf("%lld\n",sum[n]);
    }
    return 0;
}