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题目描述
神犇YY虐完数论后给傻×kAc出了一题
给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对
kAc这种傻×必然不会了,于是向你来请教……
输入格式:
第一行一个整数T 表述数据组数
接下来T行,每行两个正整数,表示N, M
输出格式:
T行,每行一个整数表示第i组数据的结果
说明
T = 10000
N, M <= 10000000
分析:题目就是求
式子化简的很好的博客
最后可以化为
后一个求和我们可以用o(n)求出前缀和,然后前面的可以整除分块求。最后时间复杂度
建议自己在纸上多推几遍
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <list>
#include <string>
#include <math.h>
#include <time.h>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, l, r) for(int i = l; i < r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#include <ext/rope>
using namespace __gnu_cxx;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;
const int N = (int) 10000000 + 11;
const int M = (int) 1e6 + 11;
const int MOD = (int) 1e9 + 7;
const double EPS = (double) 1e-9;
const double PI = (double)acos(-1.0);
const int INF = (int) 0x3f3f3f3f;
const ll INFF = (ll) 0x3f3f3f3f3f3f3f3f;
void read(int &x){
char ch = getchar(); x = 0;
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
}
/*-----------------------------------------------------------------------------------*/
bool su[N] = {1, 1, 0};
int prm[N], mu[N], g[N], sz = 0;
int sum[N];
void init(int n){
mu[1] = 1;
for(int i = 2; i <= n; i++){
if(!su[i]){
prm[++sz] = i;
mu[i] = -1;
}
for(int j = 1; j <= sz; j++){
int t = i * prm[j];
if(t > n) break;
su[t] = 1;
if(i % prm[j] == 0) {
mu[t] = 0;
break;
}else {
mu[t] = -mu[i];
}
}
}
rep(i, 1, sz + 1) for(int j = 1; j * 1ll * prm[i] <= n; j++) g[j * prm[i]] += mu[j];
sum[0] = 0;
rep(i, 1, n + 1) sum[i] = sum[i - 1] + g[i];
}
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
init(10000000);
int T;scanf("%d", &T);
while(T--){
int n, m; scanf("%d%d", &n, &m); //一开始多开了几个long long的变量居然T了几组,Emmmmm
ll ans = 0;
int c = min(n, m);
for(int l = 1, r; l <= c; l = r + 1){
r = min(n / (n / l), m / (m / l));
ans += (n / l) * 1ll * (m / l) * (sum[r] - sum[l - 1]);
}
printf("%lld\n", ans);
}
return 0;
}