HDU 4135 Co-Prime-tolerant principle

Co-Prime

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)

Problem Descriptiongiven A number N, you is asked to count the number of integers between a and B inclusive which is rel Atively Prime to N.
Integers is said to be co-prime or relatively prime if they has no common positive divisors other than 1 or, equival Ently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

Inputthe first line on input contains T (0 < T <=) the number of test cases, each of the next T lines contains t Hree integers A, B, N where (1 <= A <= B <= 1015) and (1 <=n <= 109).

Outputfor each test case, print the number of integers between A and B inclusive which is relatively prime to N. Follow T He output format below.

Sample Input21 10 23 15 5

Sample outputcase #1:5Case #2:10 HintIn the first test case, the five integers in range [1,10] which is relatively prime to 2 is {1,3,5,7,9}.

Sourcethe third Lebanese collegiate Programming contest test Instructions: give you three number, for [a, b] interval and n coprime number of numbers; thought: the number of [1,b]-[1,a-1] and n coprime because N is 1e9 , so the square root of the table, will be n only decomposition, tolerance to get the number of n not coprime; ans=b-[1,b]-(A-1-[1,a-1]);

#include <iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<Set>#include<map>using namespacestd;#definell __int64#defineMoD 1000000007#defineINF 999999999//#pragma COMMENT (linker, "/stack:102400000,102400000")intScan () {intres =0, ch;  while( ! (ch = getchar ()) >='0'&& CH <='9' ) )    {        if(ch = = EOF)return 1<< - ; } Res= CH-'0' ;  while(ch = getchar ()) >='0'&& CH <='9') Res= Res *Ten+ (CH-'0' ) ; returnRes;} ll prime[100010];ll vis[100010];ll a[ the];ll ji,cnt;ll ans,x,y,z;ll gcd (ll x,ll y) {returny==0? X:GCD (y,x%y);}voidPrime (ll N) {cnt=0; memset (Vis,0,sizeof(VIS));  for(LL i=2; i<n;i++)    {        if(!vis[i]) prime[cnt++]=i;  for(LL j=0; j<cnt&&i*prime[j]<n;j++) {vis[i*prime[j]]=1; if(i%prime[j]==0)//Key             Break; }    }}voidDFS (ll lcm,ll pos,ll step,ll X,ll &ans) {    if(lcm>x)return; if(pos==ji) {        if(step==0)            return; if(step&1) ans+ = (x/LCM); Elseans-= (x/LCM); return; } DFS (Lcm,pos+1, Step,x,ans); DFS (LCM/GCD (A[POS],LCM) *a[pos],pos+1, step+1, X,ans);}intMain () {ll t,i; intcs=1; Prime (100000); scanf ("%i64d",&t);  while(t--) {Ji=0; scanf ("%i64d%i64d%i64d",&x,&y,&z);  for(i=0; i<cnt;i++)        if(z%prime[i]==0) {A[ji++]=Prime[i];  while(z%prime[i]==0) Z/=Prime[i]; }        if(z>1) A[ji++]=Z; LL GG=0; Ans=0; DFS (1,0,0, X-1, GG); DFS (1,0,0, Y,ans); printf ("Case #%d:", cs++); printf ("%i64d\n", (Y-ans)-(x1-gg)); }    return 0;}

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HDU 4135 Co-Prime-tolerant principle