目录
一.最大周长
题目链接:https://www.lanqiao.cn/problems/21224/learning/
1.题目讲解
或者我们可以直接使用结论(长度变长,宽度小 => 周长最长)
2.代码实现
cpp
#include <iostream>
using namespace std;
int main()
{
int ret = 0;
for(int i = 1;i <= 2025;i++)
{
if(2025 % i == 0)
{
ret = max(ret,(i + 2025 / i) * 2);
}
}
cout << ret << endl;
return 0;
}二.拔河
题目链接:https://www.lanqiao.cn/problems/21223/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
#include <algorithm>
using namespace std;
#define int long long
const int N = 1 << 21;
int a[] = {0, 345590635693812, 411735179294186, 190029355501347, 973598561303630,
18202819016954, 739089526396984, 41064501651340, 287075700776565,
458062562307032, 723278851371706, 997720296178889, 470475557480472,
329586527903215, 907379737442406, 631284976214798, 301204036247736,
747294692547790, 914091289062262, 144070679727924, 988094642462741,
413975599277375, 835461430976017, 344371572186185, 646160866308904,
880407857470630, 794629069521762, 462180977651587, 342038139286302,
854772507978666, 694223418935656, 567502001946067, 881035713848915,
840605474892139, 324727089144326, 226008847101330, 65143946718125,
499249957077991, 245803813100131, 447887480320685, 658036302578844
};
int x[N],c1,y[N],c2;
void ldfs(int pos,int sum)
{
if(pos > 20)
{
x[++c1] = sum;
return;
}
ldfs(pos + 1,sum + a[pos]);
ldfs(pos + 1,sum - a[pos]);
}
void rdfs(int pos,int sum)
{
if(pos > 40)
{
y[++c2] = sum;
return;
}
rdfs(pos + 1,sum + a[pos]);
rdfs(pos + 1,sum - a[pos]);
}
signed main()
{
ldfs(1,0);
rdfs(21,0);
sort(x + 1,x + c1 + 1);
sort(y + 1,y + c2 + 1);
int ret = 1e18;
for(int i = 1,j = c2;i <= c1;i++)
{
while(j >= 2 && x[i] + y[j - 1] >= 0)
{
j--;
}
ret = min(ret,abs(x[i] + y[j]));
if(j >= 2)
{
ret = min(ret,abs(x[i] + y[j - 1]));
}
}
cout << ret << endl;
return 0;
}三.铺设能源管道
题目链接:https://www.lanqiao.cn/problems/21267/learning/
1.题目讲解
我们能看到 1 / 10 / 100 / 1000 / 10000等数的花费是最小的
2.代码实现
cpp
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int ret = 1;
while(ret < n)
{
ret *= 10;
}
cout << ret << endl;
return 0;
}四.打破规则
题目链接:https://www.lanqiao.cn/problems/21268/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
using namespace std;
int main()
{
int a,b,c;
cin >> a >> b >> c;
if(a == b || a == c || c == b || (a > b && b > c) || (a < b && b < c))
{
cout << 0 << endl;
}
else
{
cout << min(min(abs(a - b),abs(a - c)),abs(b - c)) << endl;
}
return 0;
}五.数字配对
题目链接:https://www.lanqiao.cn/problems/21211/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
#include <vector>
using namespace std;
const int N = 1e6 + 10;
int n,m;
vector<int> p[N];
int main()
{
cin >> n;
for(int i = 1;i <= n;i++)
{
int x;
cin >> x;
p[x].push_back(i);
m = max(m,x);
}
int cnt = 0;
for(int x = 1;x < m;x++)
{
if(p[x].empty() || p[x + 1].empty())
{
continue;
}
int i = p[x].size() - 1,j = p[x + 1].size() - 1;
while(i >= 0 && j >= 0)
{
if(p[x][i] < p[x + 1][j])
{
cnt++;
i--;
j--;
p[x + 1].pop_back();
}
else
{
i--;
}
}
}
cout << cnt << endl;
return 0;
}六.宗门大比
题目链接:https://www.lanqiao.cn/problems/21221/learning/
1.题目讲解
然后将其差值进行排序即可
2.代码实现
cpp
#include <iostream>
#include <algprithm>
using namespace std;
const int N = 5e5 + 10;
int n,m,k,a[N],mx;
int d[N],tot,num;
int main()
{
cin >> n >> m >> k;
for(int i = 1;i <= n;i++)
{
cin >> a[i];
mx = max(mx,a[i]);
}
if(a[k] == mx)
{
for(int i = 1;i < k;i++)
{
if(a[i] == mx)
{
num++;
}
}
cout << num + 1 << endl;
return 0;
}
for(int i = 1;i <= n;i++)
{
if(a[i] == mx)
{
num++;
continue;
}
if(a[i] >= a[k] && i < k)
{
d[++tot] = a[i] - a[k] + 1;
}
if(a[i] > a[k] && i > k)
{
d[++tot] = a[i] - a[k];
}
}
num += tot;
sort(d + 1,d + 1 + tot);
for(int i = 1;i <= tot;i++)
{
if(m >= d[i])
{
m -= d[i];
num--;
}
}
cout << num + 1 << endl;
return 0;
}七.数列染色
题目链接:https://www.lanqiao.cn/problems/21219/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int n, k, a[N];
LL f[N], q[N], h = 1, t;
int main()
{
cin >> n >> k;
for (int i = 1; i <= n; i++)
cin >> a[i];
f[1] = a[1];
for (int i = 2; i <= n; i++)
{
// 维护单调队列：队尾弹出比当前值大的元素
while (h <= t && f[q[t]] >= f[i - 1])
t--;
// 当前元素下标入队
q[++t] = i - 1;
// 队头弹出超出窗口左边界的元素
if (h <= t && q[h] < i - k - 1)
h++;
// 状态转移：取窗口内最小值 + 当前a[i]
f[i] = f[q[h]] + a[i];
}
cout << f[n] << endl;
return 0;
}八.正方形构造
题目链接:https://www.lanqiao.cn/problems/21210/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1010, mod = 1e9 + 7;
int n, m = 1000;
LL cnt[N];
// 排列数 A(n, m) 函数
LL A(LL n, LL m)
{
LL ret = 1;
for (int i = 0; i < m; i++)
ret = ret * (n - i) % mod;
return ret;
}
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
int x; cin >> x;
cnt[x]++;
}
LL ret = 0;
// 情况1：选出两个不同的数，各取2个，再排列
for (int i = 1; i <= m; i++)
{
if (cnt[i] < 2) continue;
for (int j = i + 1; j <= m; j++)
{
if (cnt[j] < 2) continue;
ret = (ret + A(cnt[i], 2) * A(cnt[j], 2) % mod) % mod;
}
}
ret = ret * 2 % mod;
// 情况2：选出相同的数，取4个排列
for (int i = 1; i <= m; i++)
{
if (cnt[i] < 4) continue;
ret = (ret + A(cnt[i], 4)) % mod;
}
cout << ret << endl;
return 0;
}九.杨辉三角
题目链接:https://www.lanqiao.cn/problems/21212/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
#include <map>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int m, cnt[N];
map<int, int> mp;
int main()
{
cin >> m;
for (LL n = 2; n <= m; n++)
{
LL c = 1;
for (LL k = 1; k <= n / 2; k++)
{
// C(n, k)
c = c * (n - k + 1) / k;
if (c > N) break; // 剪枝
if (k * 2 == n)
cnt[c] += 1;
else
cnt[c] += 2;
}
}
for (int i = 2; i <= m; i++)
mp[cnt[i]]++;
for (auto& t : mp)
cout << t.first << " " << t.second << endl;
return 0;
}十.整齐的数
题目链接:https://www.lanqiao.cn/problems/21209/learning/
1.题目讲解
2.代码实现
cpp
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int N = 20, M = 210;
LL a[N], cnt, n, m;
LL f[N][N][M];
LL dfs(int pos, int prev, int sum, bool limits, bool zero)
{
if (!pos)
return 1;
if (!limits && !zero && ~f[pos][prev][sum])
return f[pos][prev][sum];
LL ret = 0;
LL up = limits ? a[pos] : 9;
for (int i = 0; i <= up; i++)
{
if (zero)
{
ret += dfs(pos - 1, i, sum, limits && (i == a[pos]), zero && !i);
}
else if (abs(i - prev) + sum <= m)
{
ret += dfs(pos - 1, i, sum + abs(i - prev), limits && (i == a[pos]), zero && !i);
}
}
if (!limits && !zero)
f[pos][prev][sum] = ret;
return ret;
}
int main()
{
cin >> n >> m;
memset(f, -1, sizeof(f));
while (n)
{
a[++cnt] = n % 10;
n /= 10;
}
cout << dfs(cnt, 0, 0, 1, 1) << endl;
return 0;
}