Divisors UVA

打表法求出范围内的所有的素因子，然后找出区间内每一个数的素因子的个数，将这些个数相乘也就得到最终的结果，题目不难，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;int N;const int maxn = 34000;int vis[maxn + 5];vector<int> prime;void Init(){memset(vis, 0, sizeof(vis));int up = floor(sqrt(maxn) + 0.5);vis[0] = vis[1] = 1;for (int i = 2; i <= up; i++){if (!vis[i]){for (int j = i*i; j <= maxn; j+=i)vis[j] = 1;}}for (int i = 2; i <= maxn; i++){if (!vis[i]) prime.push_back(i);}}long long getRes(int data){long long sum = 1;for (int i = 0; i < prime.size() && data>1; i++){if (data%prime[i] == 0){long long amount = 1;while (data%prime[i] == 0){data = data / prime[i];amount++;}sum *= amount;}}return sum;}int main(){Init();cin >> N;while (N--){long long L, U;cin >> L >> U;long long ans = -1;int ind;for (int i = L; i <= U; i++){long long t = getRes(i);if (t > ans){ans = t;ind = i;}}cout <<"Between "<<L<<" and "<<U<<", "<<ind<<" has a maximum of "<<ans<<" divisors.\n";}return 0;}