POJ 3993 Not So Flat After All

原来就是对两个数分别因数分解，然后分别放到两个集合里，求合并后的集合的大小。再求每个数与另一个集合中该数的幂级数之差，如果另一个集合中没有这个数，就相当于这个集合中这个数的幂级数是0，最后把这些差的绝对值相加就是结果。

/* ID: sdj22251 PROG: calfflac LANG: C++ */ #include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAX 100000000 #define LOCA #define PI acos(-1.0) using namespace std; bool tag[1000011]; int p[1000011]; int cnt; void get_prime() { cnt = 0; tag[1] = 1; for (int i = 2; i < 1000005; i++) { if (!tag[i]) p[cnt++] = i; for (int j = 0; j < cnt && p[j] * i < 1000005; j++) { tag[i*p[j]] = 1; if (i % p[j] == 0) break; } } } int main() { #ifdef LOCAL freopen("ride.in","r",stdin); freopen("ride.out","w",stdout); #endif int a, b, i, cas = 0; get_prime(); while(scanf("%d%d", &a, &b) != EOF) { if(a == 0 && b == 0) break; int ans = 0; int count = 0; for(i = 0; i < cnt && (p[i] <= a || p[i] <= b); i++) { bool fg1 = false; bool fg2 = false; int ct1 = 0; int ct2 = 0; if(a % p[i] == 0) { fg1 = true; while(a % p[i] == 0) { a /= p[i]; ct1++; } } if(b % p[i] == 0) { fg2 = true; while(b % p[i] == 0) { b /= p[i]; ct2++; } } if(fg1 || fg2) { count++; ans += abs(ct1 - ct2); } } printf("%d. %d:%d\n", ++cas, count, ans); } return 0; }