FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a ，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

　　第一行包含一个正整数n，表示一共有n组询问。（1<=n<= 50000）接下来n行，每行表示一个询问，每行三个 正整数，分别为a,b,d。（1<=d<=a,b<=50000）

　　对于每组询问，输出到输出文件zap.out一个正整数，表示满足条件的整数对数。

  1 /**  2  * bzoj  3  * Problem#1101  4  * Accepted  5  * Time:8956ms  6  * Memory:1576k  7  */  8 #include <iostream>  9 #include <cstdio> 10 #include <ctime> 11 #include <cmath> 12 #include <cctype> 13 #include <cstring> 14 #include <cstdlib> 15 #include <fstream> 16 #include <sstream> 17 #include <algorithm> 18 #include <map> 19 #include <set> 20 #include <stack> 21 #include <queue> 22 #include <vector> 23 #include <list> 24 #ifndef WIN32 25 #define Auto "%lld" 26 #else 27 #define Auto "%I64d" 28 #endif 29 using namespace std; 30 typedef bool boolean; 31 const signed int inf = (signed)((1u << 31) - 1); 32 const signed long long llf = (signed long long)((1ull << 61) - 1); 33 const double eps = 1e-6; 34 const int binary_limit = 128; 35 #define smin(a, b) a = min(a, b) 36 #define smax(a, b) a = max(a, b) 37 #define max3(a, b, c) max(a, max(b, c)) 38 #define min3(a, b, c) min(a, min(b, c)) 39 template<typename T> 40 inline boolean readInteger(T& u){ 41     char x; 42     int aFlag = 1; 43     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1); 44     if(x == -1) { 45         ungetc(x, stdin);     46         return false; 47     } 48     if(x == ‘-‘){ 49         x = getchar(); 50         aFlag = -1; 51     } 52     for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘); 53     ungetc(x, stdin); 54     u *= aFlag; 55     return true; 56 } 57  58 const int limit = 5e4; 59  60 int n; 61 int a, b, d; 62 int num = 0; 63 int prime[10000]; 64 int miu[limit + 1]; 65 boolean vis[limit + 1]; 66  67 inline void Euler() { 68     memset(vis, false, sizeof(vis)); 69     miu[0] = 0, miu[1] = 1; 70     for(int i = 2; i <= limit; i++) { 71         if(!vis[i])    miu[i] = -1, prime[num++] = i; 72         for(int j = 0; j < num && prime[j] * 1LL * i <= limit; j++) { 73             int c = prime[j] * i; 74             vis[c] = true; 75             if((i % prime[j]) == 0) { 76                 miu[c] = 0; 77                 break; 78             } else { 79                 miu[c] = -1 * miu[i]; 80             } 81         } 82         miu[i] += miu[i - 1]; 83     } 84 } 85  86 inline void init() { 87     readInteger(n); 88 } 89  90 long long res; 91  92 inline void solve() { 93     while(n--) { 94         readInteger(a); 95         readInteger(b); 96         readInteger(d); 97         a /= d, b /= d; 98         if(a > b)    swap(a, b); 99         res = 0;100         for(int i = 1, j; i <= a; i = j + 1) {101             j = min(a / (a / i), b / (b / i));102             res += (a / j) * 1LL * (b / j) * (miu[j] - miu[i - 1]);103         }104         printf(Auto"\n", res);105     }106 }107 108 int main() {109     Euler();110     init();111     solve();112     return 0;113 }