hdu4135数的素数分解+容斥定理

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Co-prime

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 16   Accepted Submission(s) : 6[align=left]Problem Description[/align]Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer. 
[align=left]Input[/align]The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10[sup]15[/sup]) and (1 <=N <= 10[sup]9[/sup]). 
[align=left]Output[/align]For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below. 
[align=left]Sample Input[/align]

2
1 10 2
3 15 5
 
[align=left]Sample Output[/align]

Case #1: 5
Case #2: 10

[hint]In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}. [/hint] 
[align=left]Source[/align]The Third Lebanese Collegiate Programming Contest
题意：a~b范围内于n互质的数的个数
我们转变为找不互质的个数
找到1~b内不互质的个数和1~(a-1)内不互质的个数
答案为(b-(a-1))-(solve(b)-solve(a-1));
因此我们需要找到先n的质因数
然后根据容斥定理奇加偶减得到不互质的数的个数#include<bits/stdc++.h>
#define LL __int64
using namespace std;
int t;
LL a,b,n;
LL num[110];
LL que[100010];
LL cnt;
void divid(LL x)//分解为质因子相乘
{
cnt=0;
for(LL i=2;i*i<=x;i++)
{
if(x%i==0)
{
while(x%i==0)
{
x=x/i;
}
num[++cnt]=i;
}
}
if(x!=1)
num[++cnt]=x;
}
LL solve(LL x)
{
LL ans=0;
LL k;
LL t=0;
que[0]=-1;
for(LL i=1;i<=cnt;i++)
{
k=t;
for(LL j=0;j<=k;j++)
{
que[++t]=-1*que[j]*num[i];
}
}
for(LL i=1;i<=t;i++)
ans=ans+(x/que[i]);
return ans;
}
int main()
{
scanf("%d",&t);
int cas=0;
while(t--)
{
cas++;
scanf("%I64d%I64d%I64d",&a,&b,&n);
divid(n);
LL ans=(b-(a-1))-(solve(b)-solve(a-1));
printf("Case #%d: ",cas);
printf("%I64d\n",ans);
}
return 0;
}