标签：acm   算法   暴力   

An easy problem

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 ?

2
1
3

0
1

求满足n=i*j+i+j(0<i<=j)的i、j的种数。

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);
    }
}

HDU 2601 An easy problem(暴力枚举/质因子分解)

标签：acm   算法   暴力   

原文地址：http://blog.csdn.net/noooooorth/article/details/47115027