2301的弱化版。。。（弱过头了的说）
真是。。。为什么2301都1A了这道题却1RE+1A啊。。。蒟蒻到底了。。。
什么时候搞懂了再写题解什么的。。。
1 /************************************************************** 2 Problem: 2045 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:360 ms 7 Memory:9596 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef long long ll; 16 const int N = 1000005; 17 int u[N], p[N], tot; 18 bool v[N]; 19 20 inline int read(){ 21 int x = 0; 22 char ch = getchar(); 23 while (ch < '0' || ch > '9') 24 ch = getchar(); 25 26 while (ch >= '0' && ch <= '9'){ 27 x = x * 10 + ch - '0'; 28 ch = getchar(); 29 } 30 return x; 31 } 32 33 void pre_work(){ 34 u[1] = 1; 35 int i, j, K; 36 for (i = 2; i < N; ++i){ 37 if (!v[i]) 38 p[++tot] = i, u[i] = -1; 39 for (j = 1; i * p[j] < N && j <= tot; ++j){ 40 v[K = i * p[j]] = 1; 41 if (i % p[j] == 0){ 42 u[K] = 0; 43 break; 44 }else u[K] = -u[i]; 45 } 46 } 47 for (i = 2; i < N; ++i) 48 u[i] += u[i - 1]; 49 } 50 51 ll work(int n, int m, int k){ 52 n /= k, m /= k; 53 if (n > m) swap(n, m); 54 ll res = 0; 55 int i, last; 56 for (i = 1; i <= n; i = last + 1){ 57 last = min(n / (n / i), m / (m / i)); 58 res += (ll) (u[last] - u[i - 1]) * (n / i) * (m / i); 59 } 60 return res; 61 } 62 63 int main(){ 64 pre_work(); 65 int a = read(), b = read(), k = read(); 66 printf("%lld ", work(a, b, k)); 67 return 0; 68 }