nowcoderG 小国的复仇

题意:一开始s1=1,s2=1,每次有两种操作,1:s1=s1+s2,s2=s2 2:s1=s1*2,s2=s1,问最少几次能让S1变为n

题解:倒着推回来,可以发现n为素数,

if(n != 2*m) m = get(n-m);

n -= m;

#include<bits/stdc++.h>
#define mk(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
const int maxn = 200005, N = 1000010;
int get(int x) {
    for(ll i=2; i*i<=x; ++i) if(x%i==0) return x/i;
    return 1;
}
bool isprime[N];
ll prime[N], num;
void doprime(ll n){
    ll i,j;
    num = 0;
    memset(isprime, true,sizeof(isprime));
    isprime[1] = 0;
    for(i=2;i<=n;i++){
        if(isprime[i]){
            prime[num++] = i;
            for(j=i*i;j<=n;j+=i) isprime[j] = false;
        }
    }
}
int T, n, m, ans;
int main()
{
    doprime(1000000);
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        m = get(n);
        if(((n-1)&n) == 0&&n>10) ans = 1;
        else ans = 0;
        while(n!=1){
            if(isprime[n]){
                ans += n-1;
                break;
            }
            ++ans;
            if(n != 2*m) n -= m, m = m;
            else n /= 2, m = get(n);
        }
        printf("%d\n", ans);
    }

    return 0;
}