2016多校4 hdu 5768 Lucky7 数论+容斥原理

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题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5768

题意

在给定范围内能够整除7并且不满足模pi等于ri的数的个数。

思路

代码

#include<cstdio>#include<cstdlib>#define LL long longLL p[20],r[20];int n;LL sum;LL powMod(LL a,int b,int mod){    LL ret=1;    while(b){        if(b&1) ret=ret*a%mod;        a=a*a%mod;        b/=2;       }    return ret; }void exgcd(LL a,LL b,LL& d,LL& x,LL& y){      if(b==0){          d=a;          x=1;          y=0;      }      else{          exgcd(b,a%b,d,y,x);          y-=a/b*x;      }  }  void dfs(int i,int nu,int x,LL pp,LL rr,LL b){      //printf("%d %d %d %d %lld\n",i,nu,x,mu,b);      if(nu==x){        sum+=b/pp;        if((b%pp)&&(b%pp>=rr)) ++sum;          return;      }      if(i==n) return;    LL ppp=pp,rrr=rr,d,xx,yy;    exgcd(ppp,p[i],d,xx,yy);    xx*=(r[i]-rr)/d;    xx=(xx%(p[i]/d)+p[i]/d)%(p[i]/d);    rrr+=xx*ppp;    ppp=ppp/d*p[i];    rrr=(rrr%ppp+ppp)%ppp;      dfs(i+1,nu+1,x,ppp,rrr,b);      dfs(i+1,nu,x,pp,rr,b);      } LL rong(LL x){      LL s=0;     for(int i=1;i<=n;++i){          sum=0;          dfs(0,0,i,1,0,x);          //printf("rong%d %lld\n",i,sum);          if(i&1)              s+=sum;          else              s-=sum;      }      //printf("%lld\n",s);      return x-s;  } int main(){    int t;    scanf("%d",&t);    for(int ca=1;ca<=t;++ca){        LL x,y;        LL ansx,ansy;        scanf("%d%I64d%I64d",&n,&x,&y);        x=(x-1)/7;        y=y/7;        for(int i=0;i<n;++i){            scanf("%d%d",p+i,r+i);            r[i]=powMod(7,p[i]-2,p[i])*r[i]%p[i];           }        ansx=rong(x);        ansy=rong(y);        printf("Case #%d: %I64d\n",ca,ansy-ansx);    }      return 0;  }

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