An easy problem
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7963 Accepted Submission(s): 1920
Problem 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 <= 1010).
Output
For each case, output the number of ways in one line.
Sample Input
2 1 3
Sample Output
0 1
求满足n=i*j+i+j(0<i<=j)的i、j的种数。
第一种方法：首先这个等式可以化成(n+1)=(i+1)*(j+1),所以只要求出(n+1)的约数的种数即可。同时注意到i与j呈负相关，同时i小于等于j，所以只需要从2到sqrt(n+1)枚举（从2开始是因为i最小为1，我们枚举的是(i+1)）。但是，好暴力啊。所以我们可以用筛法先保存1e5以内的素数，再通过质因子分解求出约数数量。
第二种方法：观察等式n=i*j+i+j，可以转化成n-i=(i+1)*j，发现暴力枚举i，判断(n-i)%(n+1)==0判断整数j是否存在也是可行的，同样是因为之前说过的的原因，只需要从1到sqrt(n)枚举，同时判断i<=(n-i)/(i+1)，不符合即则跳出循环，因为题目规定i<=j。
/*
(n+1)=(i+1)*(j+1)
Memory: 1568 KB Time: 2184 MS
Language: G++ Result: Accepted
*/
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
typedef long long LL;
const double eps=1e-6;
const int MAXN=210;
LL n;
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%lld",&n);
n++;
int ans=0;
for(int i=2; (LL)i*i<=n; i++)
{
if(n%i==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}
/*
n-i=(i+1)*j
Memory: 1572 KB Time: 2059 MS
Language: G++ Result: Accepted
*/
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
typedef long long LL;
const double eps=1e-6;
const int MAXN=210;
LL n;
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%lld",&n);
if(n==1||n==2)
{
printf("0\n");
continue;
}
if(n==3)
{
printf("1\n");
continue;
}
int ans=0;
double s=sqrt(n);
for(LL i=1; i<=s; i++)
{
if(i>(n-i)/(i+1))
break;
if((n-i)%(i+1)==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}
/*
质因子分解求出约数数量
Memory: 2004 KB Time: 78 MS
Language: G++ Result: Accepted
*/
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
typedef long long LL;
const double eps=1e-6;
const int MAXN=1e5+20;
int prime[MAXN],vis[MAXN],cnt;
LL n;
int getPrime()
{
cnt=1;
vis[1]=1;
for(int i=2;i<=MAXN;i++)
{
if(vis[i])
continue;
prime[cnt++]=i;
if((LL)i*i>MAXN)
continue;
for(int j=i*i;j<MAXN;j+=i)
vis[j]=1;
}
}
int main()
{
getPrime();
int tcase;
scanf("%d\n",&tcase);
while(tcase--)
{
scanf("%lld",&n);
int ans=1;
n++;
for(int i=1;(LL)prime[i]*prime[i]<=n;i++)
{
if(n%prime[i]==0)
{
int s=0;
while(n%prime[i]==0)
{
s++;
n/=prime[i];
}
ans*=(s+1);
}
if(n==1)
break;
}
if(n!=1)
ans*=2;
printf("%d\n",(ans+1)/2-1);
}
}
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