You are in an m x n grid. You are standing in position (0, 0) and in each of the other lattice points (points with integer co-ordinates) an enemy is waiting. Now you have a ray gun that can fire up to infinity and no obstacle can stop it. Your target is to kill all the enemies. You have to find the minimum number of times you have to fire to kill all of them. For a 4 x 4 grid you have to fire 13 times. See the picture below:

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains two integers m, n (0 ≤ m, n ≤ 10⁹) and at least one of them will be less than or equal to 10⁶.

Output

For each case, print the case number and the minimum number of times you have to fire to kill all the enemies.

思路：本来想用欧拉函数的，然后一看范围太大；那么只能在[1，n]暴力每个数在[1，m]中有多少个与它互素的数，那么暴力循环[1，n]然后每次容斥找与n不互素的数，然后得互素的数。特判n=0||m=0的时候。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<math.h>
  8 #include<vector>
  9 using namespace std;
 10 typedef long long LL;
 11 bool prime[1000005];
 12 int ans[1000005];
 13 int flag[1000005];//记忆化，当一些数的质因子种类相同，避免重复运算
 14 int fen[100];
 15 int d[1000005];//每个数的最大质因数
 16 int slove(int n,int m);
 17 int main(void)
 18 {
 19     int i,j,k;
 20     fill(ans,ans+1000005,1);
 21     fill(d,d+1000005,1);
 22     for(i=2; i<=1000000; i++)
 23     {
 24         if(!prime[i])
 25         {
 26             for(j=2; (i*j)<=1000000; j++)
 27             {
 28                 prime[i*j]=true;
 29                 ans[i*j]*=i;
 30                 d[i*j]=i;
 31             }
 32         }
 33     }
 34     for(i=2; i<=1000000; i++)
 35     {
 36
 37         if(!prime[i])
 38         {
 39             ans[i]*=i;
 40             d[i]=i;
 41         }
 42     }
 43     int s;
 44     scanf("%d",&k);
 45     LL sum=0;
 46     int n,m;
 47     for(s=1; s<=k; s++)
 48     {   sum=0;
 49         memset(flag,-1,sizeof(flag));
 50         scanf("%d %d",&n,&m);
 51         if(n>m)
 52         {
 53             swap(n,m);
 54         }
 55         if(m==0)sum=0;
 56         else if(n==0)
 57         {
 58             sum=1;
 59         }
 60         else
 61         {   sum=2;
 62             for(i=1; i<=n; i++)
 63             {
 64                 if(flag[ans[i]]!=-1)
 65                 {
 66                     sum+=flag[ans[i]];
 67                 }
 68                 else
 69                 {
 70                     flag[ans[i]]=slove(i,m);
 71                     sum+=flag[ans[i]];
 72                 }
 73             }
 74         }
 75         printf("Case %d: %lld\n",s,sum);
 76     }
 77     return 0;
 78 }
 79 int slove(int n,int m)
 80 {
 81     int i,j,k;
 82     int nn=n;
 83     int cnt=0;
 84     while(n>1)
 85     {fen[cnt++]=d[n];
 86         n/=d[n];
 87     }
 88     int cc=1<<cnt;
 89     LL sum=0;
 90     int sum1=0;
 91     for(i=1; i<cc; i++)
 92     {
 93         int ck=0;
 94         int ak=1;
 95         for(j=0; j<cnt; j++)
 96         {
 97             if(i&(1<<j))
 98             {
 99                 ak*=fen[j];
100                 ck++;
101             }
102         }
103         if(ck%2)
104         {
105
106             sum+=m/ak;
107         }
108         else sum-=m/ak;
109     }
110     return m-sum;
111 }