题目大意：有n种冰淇淋和一个k面的骰子，问至少扔几次骰子使得选择每种骰子的概率相等。
先把大神的代码先放这，还没看懂
当然也可以用快速幂试试
// Time complexity: O(log(n))
// Memory: O(1)
// @EXPECTED_RESULTS@: CORRECT
/* Solution method:
*
* Insight: n divides k^t iff n divides d^t, where d = gcd(n,k).
* Hence n divides k^t iff n/d divides d^(t-1).
* Let n' = n/d, k' = d and t' = t-1. Then n divides k^t iff n' divides k'^t'.
* We can repeat this process.
* If at some point we have n' = 1, then t' = 0 is the smallest t';
* t is then the number of iterations.
* If it is not unbounded, we should eventually get t' = 0. Then n' = 1 must hold.
* Hence, if we never reach n' = 1, it is unbounded.
* We never reach n' = 1 iff at some point k' = 1 (and n' > 1).
*/
#include <algorithm>
#include <cstdio>
#include<iostream>
using namespace std;
// Greatest common divisor
int gcd (int a, int b)
{ return b ? gcd(b, a%b) : a;
}
int main()
{ int runs, run, n, k, t;
scanf("%d", &runs);
for (run = 0; run < runs; run++)
{
scanf("%d %d", &n, &k);
//cout<<"n="<<n<<",k="<<k<<endl;
for (t = 0; n > 1 && k > 1; t++)
{
//cout<<"***n="<<n<<",k="<<k<<endl;
k = gcd(n, k);
n /= k;
}
if (n > 1)
printf("unbounded\n");
else
printf("%d\n", t);
}
return 0;
}
签到题
#include<iostream>
#include<cstdio>
using namespace std;
int main(){
int t,n,ans;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
if(n>180) n-=180;
ans=(n+5)/10;
if(ans==0) ans=18;
printf("%02d\n",ans);
}
return 0;
}