HDU 2601An easy problem-素数的运用,暴力求解

id=17433" target="_blank" style="color:blue; text-decoration:none">An easy problem

Time Limit: 3000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc.. 




One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem : 



Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ? 



Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve. 

Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ? 

 

Input

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10 ¹⁰).

 

Output

For each case, output the number of ways in one line.

 

Sample Input

2
1
3 

 

Sample Output

0
1 

 对于 N as i * j + i + j (0 < i <= j) ?

能够表示为N=i*j+i+j

所以能够化作N+1=(i+1)*(j+1);

如此就有两种思路去做，一暴力枚举，从这里能够看出来i<=sqrt(N+1);

所以一个循环就能够解决，第一份代码就是。可是耗时才一点就过题目提供的3s了。怎么办，是否还有

更好的解决方式呢，有的，N+1=(i+1)*(j+1)能够知道N+1是i+1以及j+1的倍数

如此就能够转换成求解约数的个数(约数是什么,请读者自己百度了解)

N+1=a1^p1*a2^p2*a3^p3....an^pn

当中ai是代表着质数，这个的意思是不论什么大于1的数都能够转换为有限个质数因子的乘积

如此。能够用排列组合来求。第一种有p1+1选择(能够选择0...p1)另外一种有p2+1选择(能够选择0...p2)....第n种有pn+1选择(能够选择0...pn)

所以约数个数ans=(p1+1)*(p2+1)*(p3+1)*(p4+1)*....*(pn+1)(里面还有减去1和n由于他们不属于题目要求范围)

当N+1是全然平方数的话,那么除了1以及N+1本身外,唯有最中间的约数是仅仅计算了一次,其它的数都反复的计算了两次,

当N+1不是全然平方数的话,那么除了1以及N+1本身外,其它的数都反复的计算了两次

所以能够分开推断输出，也能够直接转换输出就像代码中一样,(ans+1)/2-1,当为全然平方数时,我们须要加一除二才干使正确的结果

至于减去一，就是前面的去掉1和n这两个不符合条件的数

当为不全然平方数,ans/2-1就能够了,可是为了合成一个式子,(ans+1)/2-1,是能够取代ans/2-1的

为什么，由于(4+1)/2==4/2,这是不会影响终于结果的。

/*
Author: 2486
Memory: 1416 KB		Time: 2823 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxn=1e5;
int t;
LL n;
int main() {
    scanf("%d",&t);
    while(t--) {
        scanf("%I64d",&n);
        if(n==0||n==1) {
            printf("0\n");
            continue;
        }
        int cnt=0;
        for(int i=1; i<=sqrt(n); i++) {
            if((n+1)%(i+1)==0&&(n+1)/(i+1)>=i+1)cnt++;
        }
        printf("%d\n",cnt);
    }
}

/*
Author: 2486
Memory: 1592 KB		Time: 46 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=100000+5;
LL prime[maxn];
bool vis[maxn];
int T,cnt;
LL N;
void primes() { //初始化素数列表
    cnt=0;
    for(int i=2; i<maxn; i++) {
        if(vis[i])continue;
        prime[cnt++]=i;
        for(int j=i*2; j<maxn; j+=i) {
            vis[j]=true;
        }
    }
}
void solve(LL n) {
    LL ans=1;
    for(int i=0; prime[i]*prime[i]<=n; i++) {
        if(n%prime[i]==0) {
            int s=0;
            while(n%prime[i]==0)n/=prime[i],s++;
            ans*=(s+1);
        }
        if(n==1)break;
    }
    if(n>1)ans*=2;
    printf("%I64d\n",(ans+1)/2-1);
}
int main() {
    primes();
    scanf("%d",&T);
    while(T--) {
        scanf("%I64d",&N);
        N++;
        solve(N);
    }
    return 0;
}