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[ACM] HDU 1695 GCD (容斥原理）

編輯：關於C++

GCD

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).

Source 2008 “Sunline Cup” National Invitational Contest

給定a,b,c,d,k,其中a=1,c=1，問

有多少對gcd(x,y)==k, a<=x<=b c<=y<=d
且(1,3) (3,1)這樣的算作一對,只要保證x

首先縮小范圍，另b=b/k,d=d/k
轉化成有多少對gcd(x,y)==1。
假設b 那麼將d化為 [1,b], [b+1,d]兩部分
先求 [1,b]這一部分，gcd(x,y)==1,1<=x<=b, 1<=y<=b
只要對每個數求其歐拉函數，然後累加就可以了。
再求 [b+1,d]這一部分，gcd(x,y)==1, 1<=x<=b , b+1<=y<=d
要求x,y互質，對於每一個y，如果x能夠被y的一個素因子整除，那麼gcd(x,y)肯定不等於1

所以我們先求[1,b]中有多少個數和y不互素，也就是能被y的素因子整除，然後用b減去這個數就是我們
所要求的滿足gcd(x,y)==1的數

求得時候用到了容斥原理，加上能被1個素因子整除的，減去兩個的，加上三個的，減去四個的.....

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#define rd(x) scanf("%d",&x)
#define rd2(x,y)  scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)
using namespace std;
typedef long long ll;
const int maxn=1002;
int prime[maxn+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=maxn;i++)
    {
        if(!prime[i])
            prime[++prime[0]]=i;
        for(int j=1;j<=prime[0]&&prime[j]<=maxn/i;j++)
        {
            prime[prime[j]*i]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}

int factor[100][2];
int fatcnt;
int getFactors(int x)
{
    fatcnt=0;
    int tmp=x;
    for(int i=1;prime[i]<=tmp/prime[i];i++)
    {
        factor[fatcnt][1]=0;
        if(tmp%prime[i]==0)
        {
            factor[fatcnt][0]=prime[i];
            while(tmp%prime[i]==0)
            {
                factor[fatcnt][1]++;
                tmp/=prime[i];
            }
            fatcnt++;
        }
    }
    if(tmp!=1)
    {
        factor[fatcnt][0]=tmp;
        factor[fatcnt++][1]=1;
    }
    return fatcnt;
}

const int MAXN=100010;
int euler[MAXN+1];
void getEuler()
{
    memset(euler,0,sizeof(euler));
    euler[1]=1;
    for(int i=2;i<=MAXN;i++)
    {
        if(!euler[i])
            for(int j=i;j<=MAXN;j+=i)
        {
            if(!euler[j])
                euler[j]=j;
            euler[j]=euler[j]/i*(i-1);
        }
    }
}

int t,cas=1;
int a,b,c,d,k;
long long ans;

int cal(int n,int m)//計算1~n裡面有多少個數和m互質
{
    getFactors(m);
    int ans=0;
    for(int s=1;s<(1<b||k>d)
        {
            printf("Case %d: %d\n",cas++,0);
            continue;
        }
        if(b>d)
            swap(b,d);
        b/=k,d/=k;
        ans=0;
        for(int i=1;i<=b;i++)
            ans+=euler[i];
        for(int i=b+1;i<=d;i++)
            ans+=cal(b,i);
        printf("Case %d: %I64d\n",cas++,ans);
    }
    return 0;
}