1054 - Efficient Pseudo Code

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。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<stack>
  8 using namespace std;
  9 bool prime[1000000];
 10 int ans[1000000];
 11 int mod=1e9+7;
 12 typedef  long long LL;
 13 typedef struct pp
 14 {
 15     int x;
 16     int y;
 17 } ss;
 18 ss sum[1000];
 19 stack<ss>que;
 20 LL quick(LL n,LL m);
 21 int main(void)
 22 {
 23     memset(prime,0,sizeof(prime));
 24     int i,j,k;
 25     for(i=2; i<=1000; i++)
 26     {
 27         if(!prime[i])
 28         {
 29             for(j=i; (i*j)<=1000000; j++)
 30             {
 31                 prime[i*j]=true;
 32             }
 33         }
 34     }
 35     int cnt=0;
 36     for(i=2; i<=1000000; i++)
 37     {
 38         if(!prime[i])
 39         {
 40             ans[cnt++]=i;
 41         }
 42     }
 43     scanf("%d",&k);
 44     int s;
 45     LL x,y;
 46     for(s=1; s<=k; s++)
 47     {
 48         scanf("%lld %lld",&x,&y);
 49         int tt=0;
 50         int id=0;
 51         int an=0;
 52         while(x>1&&tt<cnt)
 53         {
 54             if((LL)ans[tt]*(LL)ans[tt]>x)
 55             {   if(an!=0)
 56             {
 57                 ss dd;
 58                 dd.x=ans[tt];
 59                 dd.y=an;
 60                 que.push(dd);
 61                 an=0;
 62             }
 63                 break;
 64             }
 65             else  if(x%ans[tt]==0)
 66             {
 67                 an++;
 68                 x/=ans[tt];
 69                 if(x==1)
 70                 { ss dd;
 71                 dd.x=ans[tt];
 72                 dd.y=an;
 73                 que.push(dd);
 74                 an=0;
 75                 }
 76             }
 77             else if(x%ans[tt]!=0)
 78             {
 79                 if(an!=0)
 80                 {
 81                     ss ask;
 82                     ask.x=ans[tt];
 83                     ask.y=an;
 84                     que.push(ask);
 85                     an=0;
 86                 }
 87                 tt++;
 88             }
 89         }
 90         if(!que.empty())
 91         {
 92             ss ap=que.top();
 93             if(x!=1)
 94             {
 95                 if(x!=ap.x)
 96                 {
 97                     ss ask;
 98                     ask.x=x;
 99                     ask.y=1;
100                     que.push(ask);
101                 }
102                 else
103                 {
104                     ss dd=que.top();
105                     que.pop();
106                     dd.y+=1;
107                     que.push(dd);
108                 }
109             }
110         }
111         else
112         {
113
114             if(x!=1)
115             {
116                     ss ask;
117                     ask.x=x;
118                     ask.y=1;
119                     que.push(ask);}
120
121         }
122         int vv=0;
123         while(!que.empty())
124         {
125             sum[vv]=que.top();
126
127             que.pop();
128             vv++;
129         }
130         LL ac=1;
131         for(i=0; i<vv; i++)
132         {
133             //printf("%d %d\n",sum[i].x,sum[i].y);
134             LL q=sum[i].x-1;
135             LL ni=quick((LL)q,(LL)(mod-2));
136             LL r=(sum[i].y*y+1)%(mod-1);
137             LL p=quick((LL)sum[i].x,(LL)(r));
138             p-=1;
139             p=(p%mod+mod)%mod;
140             p=p*ni%mod;
141             ac=ac*p%mod;
142         }
143         printf("Case %d:",s);
144         printf(" %lld\n",ac%mod);
145     }
146     return 0;
147 }
148 LL  quick(LL n,LL m)
149 {
150     LL ak=1;
151     n%=mod;
152     while(m)
153     {
154         if(m&1)
155         {
156             ak=(ak%mod)*(n%mod)%mod;
157         }
158         n=(n*n)%mod;
159         m/=2;
160     }
161     return ak;
162 }