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  • [HDU 1695] GCD

    GCD

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7002    Accepted Submission(s): 2577

    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
     
    Hint
    For 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
     
    容斥定理、具体见代码
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    #define ll long long
    #define N 100000
    
    int tot;
    int prime[N+10];
    bool isprime[N+10];
    int phi[N+10];
    void prime_pri()
    {
        tot=0;
        phi[1]=1;
        memset(isprime,true,sizeof(isprime));
        isprime[0]=isprime[1]=false;
        for(int i=2;i<=N;i++)
        {
            if(isprime[i])
            {
                prime[tot++]=i;
                phi[i]=i-1;
            }
            for(int j=0;j<tot;j++)
            {
                if(i*prime[j]>N) break;
                isprime[i*prime[j]]=false;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                else
                {
                    phi[i*prime[j]]=phi[i]*(prime[j]-1);
                }
            }
        }
    }
    int fatcnt;
    int factor[N][2];
    int getfactors(int x)
    {
        fatcnt=0;
        int tmp=x;
        for(int i=0;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;
    }
    int cal(int n,int m) //求1到n中与m互质的数的个数
    {
        int tmp,cnt,ans=0;
        getfactors(m);
        for(int i=1;i<(1<<fatcnt);i++) //0表示不选择因子
        {
            cnt=0;
            tmp=1;
            for(int j=0;j<fatcnt;j++)
            {
                if(i&(1<<j))
                {
                    cnt++;
                    tmp*=factor[j][0];
                }
            }
            if(cnt&1) ans+=n/tmp;
            else ans-=n/tmp;
        }
        return n-ans;
    }
    int main()
    {
        prime_pri();
        int T,iCase=1;
        int a,b,k;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d%d%d%d",&a,&a,&b,&b,&k);
            if(k==0) //除0特判
            {
                printf("Case %d: 0
    ",iCase++);
                continue;
            }
            a/=k,b/=k;
            if(a>b) swap(a,b);
            ll ans=0;
            for(int i=1;i<=b;i++)
            {
                if(i<=a) ans+=phi[i];
                else ans+=cal(a,i);
            }
            printf("Case %d: %lld
    ",iCase++,ans);
        }
        return 0;
    }
    趁着还有梦想、将AC进行到底~~~by 452181625
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  • 原文地址:https://www.cnblogs.com/hate13/p/4461066.html
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