标签：\n std put for main 个数 ace AC cst
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 <= 1015) and (1 <=N <= 109).
题解：先求出n的所有质因数，因为n最大为1e9所以最多10个
利用二进制来模拟是否乘上某个质因数，例如有个n为2*3*5=30
1-x中共有ans个数与其互质 ans=-x/2-x/3-x/5+x/6+x/10+x/15-x/30
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long fa[20];
long long factor(long long x)
{
long long num=0;
for(long long i=2;i*i<=x;i++)
{
if(x%i==0)
{
fa[num++]=i;
while(x%i==0)x/=i;
}
}
if(x>1)fa[num++]=x;
return num;
}
long long un(long long x,long long num)
{
long long res=0;
for(long long i=0;i<(1<<num);i++)
{
long long g=1,k=0;
for(long long j=0;j<num;j++)
{
if(i&(1<<j))
g*=fa[j],k++;
}
if(k%2)
res-=x/g;
else
res+=x/g;
}
return res;
}
int main()
{
long long T,n,i=1;
long long a,b,ans;
scanf("%I64d",&T);
while(T--)
{
scanf("%I64d %I64d %I64d",&a,&b,&n);
long long num=factor(n);
ans=un(b,num)-un(a-1,num);
printf("Case #%I64d: ",i++);
printf("%I64d\n",ans);
}
return 0;
}
标签：\n std put for main 个数 ace AC cst
原文地址：https://www.cnblogs.com/carcar/p/9058317.html
