hdu-5768 Lucky7(容斥定理+中国剩余定理)
题目链接:
Lucky7
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body and sent it back to the shore. It is said that ?? used to surrounded by 7 candles when he faced a extremely difficult problem, and always solve it in seven minutes.
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi.
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi.
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.
Input
On the first line there is an integer T(T≤20) representing the number of test cases.
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<1018) on a line where n is the number of pirmes.
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi.
It is guranteed that all the pi are distinct and pi!=7.
It is also guaranteed that p1*p2*…*pn<=1018 and 0<ai<pi<=105for every i∈(1…n).
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<1018) on a line where n is the number of pirmes.
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi.
It is guranteed that all the pi are distinct and pi!=7.
It is also guaranteed that p1*p2*…*pn<=1018 and 0<ai<pi<=105for every i∈(1…n).
Output
For each test case, first output "Case #x: ",x=1,2,3...., then output the correct answer on a line.
Sample Input
2
2 1 100
3 2
5 3
0 1 100
Sample Output
Case #1: 7
Case #2: 14
题意：
问[l,r]中有多少个数%7==0且%pi!=ai;
思路：
范围太大,用容斥原理求出%7==0&&%pi==ai,的这些再加加减减;
CRT用的白书的板子;还不太会,明天来好好学学;代码参考了http://blog.csdn.net/danliwoo/article/details/52058069
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}
const LL mod=20071027;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=(1<<8)+100;
const int maxn=(1<<8);
const double eps=1e-8;
int n,flag[20];
LL L,R,m[20],a[20];
LL cal(LL x,LL y,LL mod)
{
LL s=0,base=x;
while(y)
{
if(y&1)s=(s+base)%mod;
base=(base+base)%mod;
y>>=1;
}
return s;
}
LL gao(LL x, LL r, LL p){
return (x-r)/p;
}
void exgcd(LL fa,LL fb,LL &d,LL &x,LL &y)
{
if(fb==0){d=fa;x=1;y=0;}
else
{
exgcd(fb,fa%fb,d,y,x);
y-=x*(fa/fb);
}
}
inline LL CRT()
{
LL M=1,d,y,x=0;
For(i,0,n)if(flag[i])M=M*m[i];
For(i,0,n)
{
if(!flag[i])continue;
LL w=M/m[i];
exgcd(m[i],w,d,d,y);
y=(y%M+M)%M;
x=(x+cal(cal(y,w,M),a[i],M))%M;
}
x=(x+M)%M;
LL ans=gao(R+M,x,M)-gao(M+L-1,x,M);
return ans;
}
int main()
{
int t,Case=0;
read(t);
while(t--)
{
read(n);read(L);read(R);
For(i,0,n-1)
{
read(m[i]);read(a[i]);
}
m[n]=7;a[n]=0;flag[n]=1;
int sum=(1<<n);
LL ans=0;
For(i,0,sum-1)
{
int num=0;
for(int j=0;j<n;j++)
{
if(i&(1<<j))flag[j]=1;
else flag[j]=0;
num+=flag[j];
}
if(num&1)num=-1;
else num=1;
ans=ans+num*CRT();
}
printf("Case #%d: %lld\n",++Case,ans);
}
return 0;
}
