HDU1695-GCD（数论-欧拉函数-容斥）

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5454 Accepted Submission(s): 1957



Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number
pairs.

Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.

Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

Output

For each test case, print the number of choices. Use the format in the example.

Sample Input

2
1 3 1 5 1
1 11014 1 14409 9

Sample Output

Case 1: 9
Case 2: 736427

HintFor the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

题意： 求(1,a) 和(1,b) 两个区间 公约数为k的对数的个数

思路：将a,b分别处以k,就能够转化为(1,a/k)和(1,b/k)两个区间两两互质的个数，能够先用欧拉函数求出(1,a)两两互质的个数,(a+1，b) 能够分解质因数，由于质因数的个数最多为7能够用容斥原理计算。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 10000+10;
const int maxxn = 100000+10;
typedef long long ll;
int a,b,gcd;
ll ans;
bool isPrime[maxn];
ll minDiv[maxxn],phi[maxxn],sum[maxxn];
vector<int> prime,cnt[maxxn],digit[maxxn];

void getPrime(){
prime.clear();
memset(isPrime,1,sizeof isPrime);
for(int i = 2;i < maxn; i++){
if(isPrime[i]){
prime.push_back(i);
for(int j = i*i; j < maxn; j+=i){
isPrime[j] = 0;
}
}
}
}

void getPhi(){
for(ll i = 1; i < maxxn; i++){
minDiv[i] = i;
}
for(ll i = 2; i*i < maxxn; i++){
if(minDiv[i]==i){
for(int j = i*i; j < maxxn; j += i){
minDiv[j] = i;
}
}
}
phi[1] = 1;
sum[1] = 1;
for(ll i = 2; i < maxxn; i++){
phi[i] = phi[i/minDiv[i]];
if((i/minDiv[i])%minDiv[i]==0){
phi[i] *= minDiv[i];
}else{
phi[i] *= minDiv[i]-1;
}
sum[i] = phi[i]+sum[i-1];
}
}

void getDigit(){
for(ll i = 1; i < maxxn; i++){
int x = i;
for(int j = 0; j < prime.size()&&x >= prime[j]; j++){
if(x%prime[j]==0){
digit[i].push_back(prime[j]);
int t = 0;
while(x%prime[j]==0){
t++;
x /= prime[j];
}
cnt[i].push_back(t);
}
}
if(x!=1){
digit[i].push_back(x);
cnt[i].push_back(1);
}
}
}

int main(){
getPrime();
getPhi();
getDigit();
int ncase,T=1;
cin >> ncase;
while(ncase--){
int t1,t2;
scanf("%d%d%d%d%d",&t1,&a,&t2,&b,&gcd);
if(gcd==0){
printf("Case %d: 0\n",T++,ans);
continue;
}else{
if(a > b) swap(a,b);
a /= gcd,b /= gcd;
ans = sum[a];
for(ll i = a+1; i <= b; i++){
int d = digit[i].size();
int t = 0;
vector<int> di;
for(int k = 1; k < (1<<d); k++){
di.clear();
for(int f = 0; f < d; f++){
if(k&(1<<f)){
di.push_back(digit[i][f]);
}
}
int ji = 1;
for(int f = 0; f < di.size(); f++){
ji *= di[f];
}
if(di.size()%2==0){
t -= a/ji;
}else{
t += a/ji;
}
}
ans += a-t;
}
printf("Case %d: ",T++);
cout<<ans<<endl;
}

}
return 0;
}