Nothing for Nothing( 十六)_nothing in register p-优快云博客

本文链接：https://blog.youkuaiyun.com/REfusing/article/details/98096441

本文提供了两道算法竞赛题目的详细解答，包括ANDMinimumSpanningTree和MinimalPowerofPrime的代码实现，展示了如何通过位运算和数学算法解决复杂问题。

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题目：AND Minimum Spanning Tree

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
int p[maxn];
int lowbit(int x){
    return x & (-x);
}
queue<int>q;
int main(){
    int T, n;
    for(int i = 1; i < 20; i++){
        p[(1<<i)-1] = 1;//初始化全1的情况
    }
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        int flag = 0;
        if(p[n])printf("1\n"), flag = 1;
        else printf("0\n");
        for(int i = 2; i < n; i++){
            if(i&1){
                q.push(lowbit(i+1));
            }
            else q.push(1);
        }
        if(flag)q.push(1);
        else q.push(lowbit(n+1));//取i第一位为0的位置，这这操作非常的骚我也是看别人的题解才会的
        int i = 0;
        while(!q.empty()){
            if(i != 0)printf(" ");
            printf("%d", q.front());
            q.pop();
            i++;
        }
        printf("\n");
    }
return 0;
}

题目：Minimal Power of Prime

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 1e4+10;
int pri[maxn], vis[maxn];
int cn = 1;
void Prime(){
     for(int i = 2; i <= maxn; i++){
        if(!vis[i])pri[cn++] = i;
        for(int j = 1; j < cn && i*pri[j]<maxn; j++){
            vis[pri[j]*i] = 1;
            if(i % pri[j] == 0)break;
        }
     }
}
LL check(LL x){
   return x*x*x;
}
int solve(LL n){
    LL l=10010,r=1000000;
    LL cao;
    while(l<=r){
        LL mid=(l+r)>>1;
        LL cao=check(mid);
        if(cao>n) r=mid-1;
        else if(cao<n) l=mid+1;
        else return true;
    }
    return false;
}
int main(){
    Prime();
    int T;
    long long n;
    scanf("%d", &T);
    while(T--){
        scanf("%lld", &n);
        int ans = 1000;
        for(int i = 1; i < cn; i++){
           if(n % pri[i] == 0){
             int cn = 0;
             while(n % pri[i] == 0){
                n /= pri[i];
                cn++;
             }
             ans = min(ans, cn);
           }
           if(n == 1)break;
        }
        if(n > 10009){
           LL sq1 = (LL)sqrt(n);
           LL sq2 = (LL)sqrt(sqrt(n));
           if(sq2*sq2*sq2*sq2 == n)ans = min(ans, 4);
           else if(sq1*sq1 == n)ans = min(ans, 2);
           else if(solve(n))ans = min(ans, 3);
           else ans = 1;

        }
        printf("%d\n", ans);


    }

return 0;
}