忆灬凝的博客

遍历每个竹子经历的所有高度，和他前一个竹子经历的所有高度一一比较，如果有相同的，则次数可以-1。每读入一个竹子的高度h，就计算出单独把他砍到1需要的次数和每一次的高度。

Problem - 1159http://acm.hdu.edu.cn/showproblem.php?pid=1159打印路径

Problem - 2602http://acm.hdu.edu.cn/showproblem.php?pid=2602打印选择路径：滚动数组：

Problem - 2069http://acm.hdu.edu.cn/showproblem.php?pid=2069

蓝桥杯2021年第十二届国赛真题-巧克力 - C语言网小蓝很喜欢吃巧克力，他每天都要吃一块巧克力。一天小蓝到超市想买一些巧克力。超市的货架上有很多种巧克力，每种巧克力有自己的价格、数量和剩余的保质期天数，小蓝只吃没过保质期的巧克力，请问小蓝最少花多少钱能……https://www.dotcpp.com/oj/problem2621.html...

P2026 - [蓝桥杯2022初赛] 青蛙过河 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2026二分枚举y，于是check的方法为：每一段长度为y的区间内的石头高度总和是否>=2x。如果有一段长度为y的区间不能被经过2x次，那么肯定不行。#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 +

P2045 - [蓝桥杯2022初赛] 消除游戏 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2045

P2042 - [蓝桥杯2022初赛] 特殊时间 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2042#include<bits/stdc++.h>using namespace std;typedef long long ll;int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};int checkMD(int a[]) { int

4407. 扫雷 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/description/4410/暴力。。。能过40%吧#include <bits/stdc++.h>using namespace std;typedef pair<int, int> PLL;typedef long long ll;const int N = 1e5 + 4;map<PLL, int>a;PLL c[

P2036 - [蓝桥杯2022初赛] 统计子矩阵 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2036暴力能过80%的数据。。。#include <bits/stdc++.h>typedef long long ll;using namespace std;ll a[512][512] = { 0 };ll sum[512][512] = { 0 }; // sum[i][j]表示从a[1][1]到a[i][j]的

P2037 - [蓝桥杯2022初赛] 积木画 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2037#include <bits/stdc++.h>typedef long long ll;using namespace std;const int N = 1e7 + 4;int mod = 1e9 + 7;ll dp[N] = { 0 };ll sum[N] = { 0 };int main() {

二分枚举满足条件的最小值，即二分求下界，找第一个满足条件的下标while (left < right) { ll mid = (left + right) >> 1; if (check(mid)) { right = mid; } else { left = mid + 1; }}如果mid满足条件则right = mid，不满足条件则left = mid + 1二分枚举满足条件的最大值，即二分...

P1518 - [蓝桥杯2020初赛] 回文日期 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1518#include <bits/stdc++.h>using namespace std;typedef long long ll;set<int>s;int isRun(int year) { return year % 4 == 0 && year % 100 != 0 || y

P1320 - [蓝桥杯2017初赛]方格分割 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1320#include <bits/stdc++.h>using namespace std;typedef long long ll;int ans=0;int vis[8][8] = {0};void dfs(int i, int j) { if (i <= 0 || j <= 0 || i >=

P1322 - [蓝桥杯2017初赛]包子凑数 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1322#include <bits/stdc++.h>using namespace std;typedef long long ll;int a[128]= {0};int vis[10240]= {0};int main() { int n; cin >> n; for(int i=

P1321 - [蓝桥杯2017初赛]正则问题 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1321#include <bits/stdc++.h>using namespace std;typedef long long ll;int Index = 0;string str;int kuoHao() { // 一对括号里至少有一个 | int res = 0; int temp = 0;

P1323 - [蓝桥杯2017初赛]分巧克力 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1323#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 100004;int n,k;int H[N]= {0},W[N]= {0};int check(int x) { int cnt = 0; .

P1317 - [蓝桥杯2017初赛]迷宫 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1317#include <bits/stdc++.h>using namespace std;typedef long long ll;string g[12];int vis[12][12]= {0};int res=0;int dfs(int i,int j) { if(i < 0 || j < 0

P1366 - [蓝桥杯2018初赛]倍数问题 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1366#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 100004;int a[N] = {0};vector<int>s[1024];int main() { cin.tie(0);

P1367 - [蓝桥杯2018初赛]付账问题 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1367#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 5e5 + 4;double S;int n;vector<double>ms;int main() { ios::sync_with_

P1365 - [蓝桥杯2018初赛]全球变暖 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1365#include <bits/stdc++.h>using namespace std;typedef long long ll;int N;char g[1024][1024] = { 0 };int vis[1024][1024] = { 0 };int res = 0;int cover;void dfs

P1460 - [蓝桥杯2019初赛]糖果 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1460#include <bits/stdc++.h>#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;int dp[1 << 20]; // dp[i]为得到口味为i时需要的最少数量int sg[128] = { 0 }; // sg[i

P1458 - [蓝桥杯2019初赛]外卖店优先级 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1458#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long ll;const int N = 100004;int n, m, t;multiset<int>s[N];b

P1455 - [蓝桥杯2019初赛]迷宫 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=1455#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long ll;int g[64][64] = { 0 };int vis[64][64] = { 0 };int di[] = { 1, 0

3999. 最大公约数 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/description/4002/#include<bits/stdc++.h>using namespace std;typedef long long ll;int t;ll a, m;int main() { cin >> t; while (t--) { cin >> a &gt.

3973. 无线网络 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/3976/切记范围开long long#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long ll;const int N = 1e5 + 4;int n, m;int a[N] = { 0 }, b[N] = {

PTA | 程序设计类实验辅助教学平台千名教师建设，万道高质量题目，百万用户拼题的程序设计实验辅助教学平台https://pintia.cn/problem-sets/994805046380707840/problems/994805073643683840#include <bits/stdc++.h>#define x first#define y second#define INF 0x3f3f3fusing namespace std;typedef long long

http://poj.org/problem?id=3981http://poj.org/problem?id=3981#include <iostream>#define x first#define y secondusing namespace std;typedef long long ll;int main() { // system("chcp 65001"); cin.tie(0); cout.tie(0); // freopen("

1163 -- The Trianglehttp://poj.org/problem?id=1163#include <iostream>#define x first#define y secondusing namespace std;typedef long long ll;int dp[128][128] = { 0 };int a[128][128] = { 0 };int main() { // system("chcp 65001"); ..

#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long ll;int dp[1024][1024] = { 0 };int main() { // system("chcp 65001"); cin.tie(0); cout.tie(0); freopen("C:/Users/zhaochen/Desktop/input..

#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long ll;int a[10240] = { 0 };int n;void f() { int ans = 1; int dp[10240] = { 0 }; // dp[i]是第i个数结尾的最长递增子序列的长度 dp[1] = 1; for (int i =...

#include <iostream>#include <string>#include <vector>#include <map>#include <set>#define x first#define y secondusing namespace std;typedef long long ll;const int COIN = 101;const int MONEY = 251;int type[5] = {...

力扣https://leetcode-cn.com/problems/car-pooling/class Solution {public: bool carPooling(vector<vector<int>>& trips, int capacity) { int delta[1024] = { 0 }; // delta[j]保存第j站变化的人数 for (int i = 0; i < trips.size(); i

107. 超快速排序 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/description/109/#include <iostream>using namespace std;typedef long long ll;const int N = 1024000;ll a[N] = { 0 }, t[N] = { 0 }, cnt = 0;void Merge(ll start, ll mid, ll end)

3468 -- A Simple Problem with Integershttp://poj.org/problem?id=3468#include <iostream>#include <string>#include <vector>#include <map>#include <set>#define x first#define y secondusing namespace std;typedef long lo

链接：2182 -- Lost Cowshttp://poj.org/problem?id=2182纯暴力：#include <iostream>#define x first#define y secondusing namespace std;typedef long long ll;const int N = 8004;int ans[N], a[N], pre[N];int main() { // system("chcp 65001"); ci.

链接：3134 -- Power CalculusDescriptionStarting withxand repeatedly multiplying byx, we can computex31with thirty multiplications:x2=x×x,x3=x2×x,x4=x3×x, …,x31=x30×x.The operation of squaring can be appreciably shorten the...

单点时限：1.0 s内存限制：512 MB题目描述给出一个有个点的完全图，点的编号为，编号为的点之间的边权为。请求出这张图的最小生成树的代价。完全图：在图论的数学领域，完全图是一个简单的无向图，其中每对不同的顶点之间都恰连有一条边相连。：与的最小公倍数。最小生成树：一个有个结点的连通图的生成树是原图的极小连通子图，且包含原图中的所有个结点，并且有保持图连通的最少的边。最小生成树其实是最小权重生成树的简称。输入格式第一行一个整数，表示有...

单点时限：1.0 s内存限制：512 MB题目描述有个点，编号为，在秒时 c4Lnn 位于点，每秒 c4Lnn 都必须跳到其它点，即 c4Lnn 不能在同一点连续停留秒。求经过秒后，c4Lnn 停留在点的不同方案数。由于答案会很大，请输出答案对取模的结果。输入格式一行四个整数。输出格式一行，一个整数，表示不同方案数。样例input3133output3有以下种方案：#inclu...

Given a sequence ofKintegers {N1​,N2​, ...,NK​}. A continuous subsequence is defined to be {Ni​,Ni+1​, ...,Nj​} where1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given seq...