Divisors UVA - 294

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;
}