1054

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

---

PROBLEM SETTER: JANE ALAM JAN

思路：先打表素数，然后我们先将每个x分解成各个质因数的和，然后我们知道sum=（p1⁰+p1¹+...p1^r1）*(p2⁰+p2¹+....p2^r2)*...

pi为质因数，ri为每个质因数的个数乘以y，那么每项用等比数列求和公式求下，然后（1-q^r+1）/（1-q）;然后费马小定理求下逆元即可。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
using namespace std;
bool prime[1000000];
int ans[1000000];
int mod=1e9+7;
typedef  long long LL;
typedef struct pp
{
    int x;
    int y;
} ss;
ss sum[1000];
stack<ss>que;
LL quick(LL n,LL m);
int main(void)
{
    memset(prime,0,sizeof(prime));
    int i,j,k;
    for(i=2; i<=1000; i++)
    {
        if(!prime[i])
        {
            for(j=i; (i*j)<=1000000; j++)
            {
                prime[i*j]=true;
            }
        }
    }
    int cnt=0;
    for(i=2; i<=1000000; i++)
    {
        if(!prime[i])
        {
            ans[cnt++]=i;
        }
    }
    scanf("%d",&k);
    int s;
    LL x,y;
    for(s=1; s<=k; s++)
    {
        scanf("%lld %lld",&x,&y);
        int tt=0;
        int id=0;
        int an=0;
        while(x>1&&tt<cnt)
        {
            if((LL)ans[tt]*(LL)ans[tt]>x)
            {   if(an!=0)
            {
                ss dd;
                dd.x=ans[tt];
                dd.y=an;
                que.push(dd);
                an=0;
            }
                break;
            }
            else  if(x%ans[tt]==0)
            {
                an++;
                x/=ans[tt];
                if(x==1)
                { ss dd;
                dd.x=ans[tt];
                dd.y=an;
                que.push(dd);
                an=0;
                }
            }
            else if(x%ans[tt]!=0)
            {
                if(an!=0)
                {
                    ss ask;
                    ask.x=ans[tt];
                    ask.y=an;
                    que.push(ask);
                    an=0;
                }
                tt++;
            }
        }
        if(!que.empty())
        {
            ss ap=que.top();
            if(x!=1)
            {
                if(x!=ap.x)
                {
                    ss ask;
                    ask.x=x;
                    ask.y=1;
                    que.push(ask);
                }
                else
                {
                    ss dd=que.top();
                    que.pop();
                    dd.y+=1;
                    que.push(dd);
                }
            }
        }
        else
        {

            if(x!=1)
            {
                    ss ask;
                    ask.x=x;
                    ask.y=1;
                    que.push(ask);}

        }
        int vv=0;
        while(!que.empty())
        {
            sum[vv]=que.top();

            que.pop();
            vv++;
        }
        LL ac=1;
        for(i=0; i<vv; i++)
        {
            //printf("%d %d
",sum[i].x,sum[i].y);
            LL q=sum[i].x-1;
            LL ni=quick((LL)q,(LL)(mod-2));
            LL r=(sum[i].y*y+1)%(mod-1);
            LL p=quick((LL)sum[i].x,(LL)(r));
            p-=1;
            p=(p%mod+mod)%mod;
            p=p*ni%mod;
            ac=ac*p%mod;
        }
        printf("Case %d:",s);
        printf(" %lld
",ac%mod);
    }
    return 0;
}
LL  quick(LL n,LL m)
{
    LL ak=1;
    n%=mod;
    while(m)
    {
        if(m&1)
        {
            ak=(ak%mod)*(n%mod)%mod;
        }
        n=(n*n)%mod;
        m/=2;
    }
    return ak;
}