杭电多校（四）2019.7.31--暑假集训
作者：互联网
【HDU 6014】
SOLVED
【题目大意】给定N个节点，两点之间距离是节点编号的与，在这样的前提下，求最小生成树，输出代价和路径
【思路】通过lowbit求第一个0的位置，然后令此位为1的值就是最优解
【总结】1.与或非都要先考虑拆分后二进制的特性
2.检验算法正确性时，验证数据要是自己验证能力的最大值
（就是多验）
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<string>
#include<cstring>
#include<climits>
#include<cmath>
#include<map>
#include<set>
#include<deque>
using namespace std;
const int maxn = 2e5 + 10;
int arr[maxn];
int lowbit(int x)
{
return x & -x;
}
int main()
{
ios_base::sync_with_stdio(false);
int T;
cin >> T;
while (T--)
{
int N;
cin >> N;
int cal = 0;
for (int i = 2; i <= N; i++)
{
if ((i & 1) == 0)
arr[i] = 1;
else
{
int t =lowbit((i + lowbit(i)));
if (t <= N)
arr[i] = t;
else
{
arr[i] = 1;
cal++;
}
}
}
cout << cal << "\n";
for (int i = 2; i <= N; i++)
{
cout << arr[i];
if (i != N)
cout << ' ';
}
cout << "\n";
}
}
View Code
【HDU 6015】
UNSOLVED
【HDU 6016】
UNSOLVED
【HDU 6017】
UNSOLVED
【HDU 6018】
UNSOLVED
【HDU 6019】
UNSOLVED
【HDU 6020】
SOLVED
【题目大意】给一个4*4的数字谜题，能否恢复到指定形状
【思路】数字谜题的结论
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<string>
#include<cstring>
#include<climits>
#include<cmath>
#include<map>
#include<set>
#include<deque>
#include<unordered_map>
using namespace std;
struct node
{
int arr[5][5];
bool operator<(const node& a)const
{
for (int i = 1; i <= 4; i++)
{
for (int j = 1; j <= 4; j++)
{
if (arr[i][j] >= a.arr[i][j])
return false;
}
}
return true;
}
node() {};
node(const int a[][5])
{
for (int i = 1; i <= 5; i++)
{
for (int j = 1; j <= 5; j++)
{
arr[i][j] = a[i][j];
}
}
}
};
map<node, int>mp;
int arr[16];
const int mx[4] = { 0,-1,0,1 };
const int my[4] = { 1,0,-1,0 };
int main()
{
ios_base::sync_with_stdio(false);
int T;
cin >> T;
while (T--)
{
int cal = 0;
int px;
for (int i = 1; i <= 16; i++)
{
int n;
cin >> n;
if (n == 0)
{
px = (i - 1) / 4 + 1;
continue;
}
arr[++cal] = n;
}
int cnt = 0;
for (int i = 1; i <= 15; i++)
{
for (int j = i + 1; j <= 15; j++)
{
if (arr[i] > arr[j])
cnt++;
}
}
if ((cnt + 4 - px) % 2 == 0)
cout << "Yes\n";
else
cout << "No\n";
}
}
View Code
【HDU 6021】
UNSOLVED
【HDU 6022】
UNSOLVED
【HDU 6023】
SOLVED
【题目大意】
【思路】对素数进行特殊处理，先求N^1/5，于是剩下的素数的幂次最高只能是4了，可以二分开根迅速求解
#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
ll pri[10001];
int vis[10001];
int cntp;
void init()
{
for (ll i = 2; i < 10000; i++)
{
if (!vis[i])
{
cntp++;
pri[cntp] = i;
}
for (int j = 1; j <= cntp&&pri[j]*i<10000; j++)
{
vis[i*pri[j]] = 1;
if (i%pri[j] == 0)
{
break;
}
}
}
}
ll sqrt(ll num,int d)
{
ll l = 1, r = num;
while (l <= r)
{
ll mid = (l+r) / 2;
ll tmp = LLONG_MAX;
if(d==3)
{
tmp = tmp / mid;
}
if (d == 4)
{
tmp = tmp / mid;
tmp /= mid;
}
if (tmp / mid < mid)
{
r = mid - 1;
continue;
}
tmp = 1;
for (int i = 1; i <= d; i++)
tmp *= mid;
if (tmp == num)
return mid;
if (tmp > num)
r = mid - 1;
else
l = mid + 1;
}
return 0;
}
int main()
{
init();
int T;
scanf("%d", &T);
while (T--)
{
ll n;
scanf("%lld", &n);
int ans = INT_MAX;
for (int i = 1; i <= cntp; i++)
{
if (pri[i] > n)
break;
int cnt = 0;
while (n%pri[i] == 0)
{
cnt++;
n/= pri[i];
}
//if (cnt)
// printf("cnt %d pri %lld\n", cnt, pri[i]);
if (cnt < ans && cnt)
ans = cnt;
}
int res = INT_MAX;
if (n>1)
{
if (sqrt(n, 4))
res = min(res, 4);
else
{
if (sqrt(n, 2))
res = min(res, 2);
}
if (sqrt(n, 3))
res = min(res, 3);
if (res == INT_MAX && n > 4500)
res = 1;
}
printf("%d\n", min(ans,res));
} //printf("1000000000000000000 %lld %lld\n", sqrt(1000000000000000000, 3), sqrt(1000000000000000000, 2));
return 0;
}
View Code
标签：杭电多校,cnt,int,res,31,2019.7,HDU,UNSOLVED,include 来源： https://www.cnblogs.com/rentu/p/11297739.html