1907. Coffee and Buns
Time limit: 1.0 second
Memory limit: 64 MB
Planet Ataraxia is known for its education centers. The people who are expected to take high social positions in future are brought up in conditions of continuous training and supervision from the early
age. Each education center is a small town with all the necessary utilities. During the construction of a center, a square area is chosen which is divided into equal sections each sized 100 × 100 meters. In each of these sections they build one building, which
would become residential or academic. The outer perimeter of the center is fenced.
After a successful military operation in the Andromeda nebula the active colonization of habitable planets has started. The need for people able to take command and lead people to the new worlds has
increased. Therefore, two new education centers should be built on Ataraxia. Discussion about the details of the project in the local administration is already underway for many days. During this time it was decided that the first center will consist of a2 sections
and that the second one will consist of no more than n2 sections. The situation is complicated because, according to requirements of the antimonopoly legislation, construction
works must be performed by at least two different companies, each of them must build an equal number of buildings and an equal number of 100-meters segments of the fence.
You are responsible for resupplying the administration office. You understand that while they are discussing the pros and cons of each possible size of the second center a lot of buns and coffee will
be consumed, and it‘s time to buy them. So you‘d like to know how many different sizes of the second center will meet the requirements of antimonopoly legislation and, therefore, will be fully considered by the administration.
Input
The only line contains integers a and n (1 ≤ a ≤ 1012; 1 ≤ n ≤ 1018).
Output
Output an amount of different sizes of the second center meeting the requirements of antimonopoly legislation.
Sample
| input | output |
|---|---|
3 6 |
4 |
Notes
In this example it is possible to build the second center sized 3 × 3 or 6 × 6, delegating the construction to three different companies, or to build it sized 1 × 1 or 5 × 5, delegating the construction
to two different companies.
Problem Author: Ivan Burmistrov
Problem Source: Ural Championship 2012
Tags: none (
hide tags for unsolved problems
)
Difficulty: 985 Printable version Submit solution Discussion
(0)
All submissions (654) All accepted submissions (223) Solutions
rating (123)
大致题意：
输入 a n 判断 从 b = 1 ~ n 中 满足 gcd( ( a * a + b * b ) , 4 * a + 4 * b ) > 1 的个数。
a < 1e12 , n < 1e18
若 a b 奇偶性相同，显然满足。gcd是2的倍数。
若a b 奇偶性不同，
开始推导：
因为 a * a + b * b 是奇数，
所以
gcd( a * a + b * b , 4 * a + 4 * b ) > 1 等价于
gcd( a * a + b * b , a + b ) > 1 等价于
根据整除性质，若存在 e > 1 ，满足 e | ( a + b ) ， 那么 e | ( a + b ) ^ 2 反之亦然
则等价于
gcd( a * a + b * b , a * a + b * b + 2 * a * b ) > 1 等价于
gcd( 2 * a * a + 2 * b * b , a * a + b * b + 2 * a * b ) > 1等价于
gcd( a * a + b * b - 2 * a * b , a * a + b * b + 2 * a * b ) > 1 等价于
gcd( a - b , a + b ) > 1 等价于
gcd( 2 * a , a + b ) > 1
因为 a+b是奇数
等价于
gcd( a , a +b ) > 1等价于
gcd( a , b ) > 1
推导结束
下面对a分解质因数，找出n个数中不同奇偶性且满足gcd( a , b ) > 1的个数
最后加上 相同奇偶性的 个数 就是答案。
所以要求1~maxn中与a，gcd > 1 的个数，就是求1~maxn与某一个num不互素的个数，要用到容斥原理：
ll no_coprime(ll num,ll MAXN){//1~maxn与num不互素的个数
ll ans = 0;
vector<ll> fac;
for(ll i = 2; i*i <= num; i++){ //分解因数
if( num % i == 0){
fac.push_back(i);
while(num%i == 0) num /= i;
}
}
if(num != 1) fac.push_back(num);
int sz = SZ(fac);
for(ll mask = 1 ; mask < (1LL<<sz); mask++){ //容斥过程，复杂度2^因数个数
ll c = 0;
ll tmp = 1;
rep(i,sz) if( (1LL<<i)&mask){
c++;
tmp *= fac[i];
}
if(c&1) ans += MAXN/tmp;
else ans -= MAXN/tmp;
}
return ans;
}
ac代码：
//Accepted 412 15 G++ 4.9 C++11 2092
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
#define rep(i,n) for ( int i=0; i<int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<ll,ll> pii;
template <class T>
inline bool RD(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void PT(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
ll a,n;
ll no_coprime(ll num,ll MAXN){
ll ans = 0;
vector<ll> fac;
for(ll i = 2; i*i <= num; i++){ //分解因数
if( num % i == 0){
fac.push_back(i);
while(num%i == 0) num /= i;
}
}
if(num != 1) fac.push_back(num);
int sz = SZ(fac);
for(ll mask = 1 ; mask < (1LL<<sz); mask++){ //容斥过程，复杂度2^因数个数
ll c = 0;
ll tmp = 1;
rep(i,sz) if( (1LL<<i)&mask){
c++;
tmp *= fac[i];
}
if(c&1) ans += MAXN/tmp;
else ans -= MAXN/tmp;
}
return ans;
}
int main(){
while(cin>>a>>n){
ll ans = 0;
if(a&1) ans += (n+1)/2, n /= 2;
ans += no_coprime(a,n);
printf("%lld\n",ans);
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。