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  • HDU 5768 Lucky7

    中国剩余定理+容斥原理。

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<iostream>
    using namespace std;
    typedef long long LL;
    const double pi=acos(-1.0),eps=1e-8;
    void File()
    {
        freopen("D:\in.txt","r",stdin);
        freopen("D:\out.txt","w",stdout);
    }
    inline int read()
    {
        char c = getchar();  while(!isdigit(c)) c = getchar();
        int x = 0;
        while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
        return x;
    }
    
    const int maxn=20;
    int T,p,cas=1,limit;
    LL L,R,A[maxn],B[maxn],h,a[maxn],b[maxn];
    
    LL Extend_Euclid(LL a, LL b, LL&x, LL& y)
    {
        if(b==0)
        {
            x=1,y=0;
            return a;
        }
        LL d = Extend_Euclid(b,a%b,x,y);
        LL t = x;
        x = y;
        y = t - a/b*y;
        return d;
    }
    
    LL China_Reminder(int len, LL* a, LL* n)
    {
        int i;
        LL N = 1;
        LL result = 0;
        for(i = 0; i < len; i++) N = N*n[i];
        for(i = 0; i < len; i++)
        {
            LL m = N/n[i];
            LL x,y;
            Extend_Euclid(m,n[i],x,y);
            x = (x%n[i]+n[i])%n[i];
            result = (result + m*a[i]*x%N)%N;
        }
        return result;
    }
    
    void dfs(int x,int s)
    {
        if(s==limit)
        {
            a[s]=7, b[s]=0;
            LL res=China_Reminder(s+1,b,a); if(res>R) return;
            LL rep=1; for(int i=0;i<=s;i++) rep*=a[i];
            LL small,large;
            if(res>=L) small=0; else small=(L-res)%rep==0 ? ((L-res)/rep):((L-res)/rep+1);
            large=(R-res)/rep;
            h=h+large-small+1;
            return;
        }
        if(x==p+1) return;
        dfs(x+1,s);
        a[s]=A[x], b[s]=B[x]; dfs(x+1,s+1);
    }
    
    int main()
    {
        scanf("%d",&T); while(T--)
        {
            scanf("%d%lld%lld",&p,&L,&R);
            for(int i=1;i<=p;i++) scanf("%lld%lld",&A[i],&B[i]);
            LL sum=0,f=1;
            for(int i=1;i<=p;i++) limit=i, h=0, dfs(1,0), sum=sum+f*h, f=-f;
            printf("Case #%d: %lld
    ",cas++,R/7-(L-1)/7-sum);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5720989.html
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