"Chinese remainder theorem" "Tolerant principle" "fast multiplication" "Number theory" HDU 5768 Lucky7

Topic Links:

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

Main Topic :

T set of data, the l~r satisfies: 1. is a multiple of 7, 2. The number of%pi!=ai for n primes.

Topic Ideas:

"Chinese remainder theorem" "Tolerant principle" "fast multiplication" "Number theory"

　Because they are prime numbers, they are 22 of each other, satisfying the conditions of the Chinese remainder theorem.

The 7 is added to the prime number, a=0, so as to solve the n+1 equation of the general solution (the smallest solution). Then calculate how many solutions are in l~r.

But because the condition of the Chinese remainder theorem is established at the same time, and the topic is or the relationship, so we should use the principle of tolerance to overlay the limitation.

The middle process may explode long, so use fast multiplication (similar to fast power, just multiply to add)

1 //2 //by Coolxxx3 //4#include <iostream>5#include <algorithm>6#include <string>7#include <iomanip>8#include <memory.h>9#include <time.h>Ten#include <stdio.h> One#include <stdlib.h> A#include <string.h> - //#include <stdbool.h> -#include <math.h> the #defineMin (a) < (b) ( A):(B)) - #defineMax (a) (a) > (b)? ( A):(B)) - #defineABS (a) ((a) >0? ( A):(-(a))) - #defineLowbit (a) (a& (a)) + #defineSqr (a) ((a) * (a)) - #defineSwap (a) (a) ^= (b), (b) ^= (a), (a) ^= (b)) + #defineEPS (1E-8) A #defineJ 10000000 at #defineMAX 0x7f7f7f7f - #definePI 3.1415926535897 - #defineN 24 - using namespacestd; -typedefLong LongLL; - intCas,cass; in intn,m,lll; - LL L,r,ans; to LL P[n],a[n]; + BOOLU[n]; - ll Cheng (ll a,ll b,ll MoD) the { *LL sum=0; $      for(; b;b>>=1)Panax Notoginseng     { -         if(B&AMP;1LL) sum= (sum+a)%MoD; theA= (a+a)%MoD; +     } A     returnsum; the } +ll EXGCD (ll a,ll b,ll &x,ll &y) - { $     if(!B) {x=1, y=0;returnA;} $LL D=EXGCD (b,a%b,y,x); -y-=a/b*x; -     returnD; the } -LL China (intnn)Wuyi { theLL sum=0, tot=1, Tott,x,y; -     inti; Wu      for(i=1; i<=nn;i++)if(U[i]) tot*=P[i]; -      for(i=1; i<=nn;i++) About     { $         if(!u[i])Continue; -tott=tot/P[i]; - EXGCD (tott,p[i],x,y); -X= (X%p[i]+p[i])%P[i]; ASum= ((Sum+cheng (A[i]*tott%tot,x,tot))%tot+tot)%tot; +     } theSum= (r+tot-sum)/tot-(l1+tot-sum)/tot; -     returnsum; $ } the intMain () the { the #ifndef Online_judge the //freopen ("1.txt", "R", stdin); - //freopen ("2.txt", "w", stdout); in     #endif the     intI,j,k,ii; the //for (scanf ("%d", &cas); cas;cas--) About      for(SCANF ("%d", &cas), cass=1; cass<=cas;cass++) the //while (~scanf ("%s", s)) the //while (~scanf ("%d", &n)) the     { +ans=0; -printf"Case #%d:", Cass); thescanf"%d%lld%lld",&n,&l,&R);Bayi          for(i=1; i<=n;i++) thescanf"%lld%lld",&p[i],&a[i]); theLll=1<<N; -n++; -u[n]=1;p [n]=7; a[n]=0; the          for(i=0; i<lll;i++) the         { the              for(j=i,k=0, ii=1; ii<n;j>>=1, ii++) the             { -u[ii]=j&1; thek+=U[ii]; the             } thek=k&1?-1:1;94ans+=1ll*k*China (n); the         } theprintf"%lld\n", ans); the     }98     return 0; About } - /*101 //102 103 //104 */

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"Chinese remainder theorem" "Tolerant principle" "fast multiplication" "Number theory" HDU 5768 Lucky7