LightOJ - 1054 - Efficient Pseudo Code

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1054 - Efficient Pseudo Code

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Time Limit: 1 second(s)

Memory Limit: 32 MB

Sometimes it's quite useful to write pseudo codes for problems. Actually you can write the necessary steps to solve a particular problem. In this problem you are given a pseudo code to solve a problem and you have to implement the pseudo code efficiently. Simple! Isn't it? :)

pseudo code

{

    take two integers n and m

    let p = n ^ m (n to the power m)

    let sum = summation of all the divisors of p

    let result = sum MODULO 1000,000,007

}

Now given n and m you have to find the desired result from the pseudo code. For example if n = 12 and m = 2. Then if we follow the pseudo code, we get

pseudo code

{

    take two integers n and m

    so, n = 12 and m = 2

    let p = n ^ m (n to the power m)

    so, p = 144

    let sum = summation of all the divisors of p

    so, sum = 403, since the divisors of p are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144

    let result = sum MODULO 1000,000,007

    so, result = 403

}

Input

Input starts with an integer T (≤ 5000), denoting the number of test cases.

Each test case will contain two integers, n (1 ≤ n) and m (0 ≤ m). Each of n and m will be fit into a 32 bit signed integer.

Output

For each case of input you have to print the case number and the result according to the pseudo code.

Sample Input	Output for Sample Input
3 12 2 12 1 36 2	Case 1: 403 Case 2: 28 Case 3: 3751

　　这一题的题意很简单，就是给你2个数n,m求n的m次幂的因子和是多少，当然，结果要模1000000007。这一题的要求其实有两个，一是不超时，二是得到正确的答案，则用普通的pow函数配合求因子明显不行。

　　首先，因为数字很大，如果先pow在求模明显早溢出了，其次，太大的数字求因子和明显慢得像蜗牛。所以这一题需要自己写一个快速幂，同时要知道一件事就是我们不需要一开始就算出pow(n,m)出来，由素数唯一分解定理可以得到，pow(n,m)的分解出来的素数种类和n分解出来的种类是一样的只是数目上是n的各个素数的数目的m倍，所以我们可以先求出n的素数，然后每一个的含有数目都乘上m即可，然后就是求因子和了。求因子和需要求每一个素数的p^αi这一个等比数列，然后根据乘法原理求出结果，就是因子和这里可以很好地理解，每一个素数出现有0~p^αi这么多种状态，它们和其他素数的不同状态组合起来就得到所有的结果了。这里有很多方法，师兄的代码是用逆元来解，但是暂时我还不懂逆元怎样用，而且师兄说过这种方法有局限，因此这里我使用的是另一种解法，就是二分求等比数列。分析如下图

 

　　代码可以对两种情况进行完全分开的写求值过程，而不是偶数时前半部分化为奇数情况调用，前者大概会快一点。(= =反正两种我都试过了)

上代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#define MOD 1000000007
#define LL long long
#define MAX (1<<16)+10
using namespace std;

LL pri[MAX];
int tot;
bool mark[MAX];

void dedeal()
{
    LL i,n,j;
    n=MAX-10;
    memset(pri,0,sizeof(pri));
    memset(mark,0,sizeof(mark));
    for(i=2;i<=n;i++)
    {
        if(!mark[i])
        {
            pri[tot++]=i;
            for(j=i*i;j<=n;j+=i)
            {
                mark[j]=1;
            }
        }
    }
}

LL FastMod(LL a,LL q)
{
    if(q==0) return 1;
    LL t=FastMod(a,q>>1);
    t=t%MOD;
    t=t*t%MOD;
    if(q&1) t=t*a%MOD;
    return t;
}

/*
LL FastMod(LL a,LL q)
{
    LL k,r;
    if(!q) return 1;
    k=1;
    r=a%MOD;
    while(q>1)
    {
        if(q&1) k=k*r%MOD;
        r=r*r%MOD;
        q>>=1;
    }
    return r*k%MOD;
}
*/
LL f(LL a,LL q)
{
    if(q==0) return 1;
    if(q&1) return (f(a,q>>1)%MOD*(1+FastMod(a,(q+1)>>1)))%MOD;
    return (f(a,q-1)%MOD+FastMod(a,q))%MOD;
}

LL fs(LL n,LL M)
{
    LL i,m,sum,amo;
    i=0;
    m=sqrt(n);
    sum=1;
    for(i=0;i<tot && pri[i]<=m;i++)
    {
        if(n%pri[i]==0)
        {
            amo=0;
            //printf("%lld ",pri[i]);
            while(n%pri[i]==0) {n/=pri[i];amo++;}
            sum=sum*f(pri[i],(amo*M))%MOD;
        }
        if(n==1) break;
    }
    if(n>1) sum=sum*f(n,M)%MOD;
    return sum%MOD;
}

int main()
{
    int t,i;
    LL n,m,sum;
    //freopen("data.txt","r",stdin);
    dedeal();
    scanf("%d",&t);
    for(i=1;i<=t;i++)
    {
        scanf("%lld %lld",&n,&m);
        sum=fs(n,m);
        printf("Case %d: %lld\n",i,sum%MOD);
    }
    return 0;
}