文章分类 - OI题目 / 弱智难度
摘要:/* https://www.luogu.com.cn/problem/P1816 */ #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> //#include<queue> //#include<vect
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posted @ 2023-10-22 14:39
JMXZ
摘要:/* 最小生成树的 sum唯一 但形状不一定唯一(边长相同) 若边长唯一 则 tree 唯一 */ #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> //#include<queue> //#include
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posted @ 2023-10-01 01:50
JMXZ
摘要:RMQ 模板题 /* */ #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> //#include<queue> //#include<vector> #include<bits/stdc++.h> #de
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posted @ 2023-09-30 01:35
JMXZ
摘要:求出重心+节点到重心距离 /* Çó³öÊ÷µÄÖØÐÄ + ¼ÆËã³öÿ¸ö½Úµãµ½ ÖØÐĵľàÀë */ #include<cstdio> #include<iostream>
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posted @ 2023-09-29 08:22
JMXZ
摘要:#include<iostream> #include<cmath> using namespace std; int a; void fff(int x) { for(int i=14;i>=0;i--) //两万的数据最多是2(14) { if(pow(2,i)<=x) { //pow(n,m)
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posted @ 2023-08-22 14:18
JMXZ
摘要:#include<iostream> #include<string> #include<algorithm> using namespace std; string s[21];int n; bool cmp(string a,string b) { // &表示引用 return (a+b >=
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posted @ 2023-08-22 12:56
JMXZ
摘要:#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<vector> #include<bits/stdc++.h> #define ll long long #define ddd prin
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posted @ 2023-08-21 14:06
JMXZ
摘要://if-> else if没有逻辑 /* 5 1 100 60 70 27 88 1 100 75 80 39 34 1 100 50 605 563 50 1 200 160 120 176 40 1 200 10 50 47 84 No Yes Yes Yes No */ #include<c
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posted @ 2023-08-20 16:32
JMXZ
摘要::::easy->easy #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define ddd print
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posted @ 2023-08-13 13:23
JMXZ
摘要::::连边时 sm->big 单向边 #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<algorithm> #include<queue> using namespace std; bool
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posted @ 2023-08-10 17:17
JMXZ
摘要:>>> /3 -> 遍历一遍就可以了 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define ddd
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posted @ 2023-08-09 23:19
JMXZ
摘要:>>> de[]->big lian >>>co[] -> point belong >>>只能统计 极值 cd rd ==0 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bit
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posted @ 2023-08-07 17:19
JMXZ
摘要:>>>在有向图G中,如果两个顶点间至少存在一条路径,称两个顶点强连通。如果有向图G的每两个顶点都强连通,称G是一个强连通图。有向图非强连通图的极大强连通子图,称为强连通分量。 说白了就是如果一个有向图的子图里每个点可以两两互达,那么这个子图是一个强联通分量 -> 两两可以到达 Tarjan算法就是基
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posted @ 2023-08-07 16:56
JMXZ
摘要:>>>x+z/2 -> x z same 奇偶 >>>∑ i*(s[c[i]][i%2]*x[i]+ sum[ c[i] ][i%2] ) #include <cmath> #include <ctime> #include <cstdio> #include <string> #include <
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posted @ 2023-08-07 13:13
JMXZ
摘要::::素都会根据元素的键值自动排序 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define ddd p
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posted @ 2023-08-05 11:08
JMXZ
摘要:#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define ddd printf(" debug\n");
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posted @ 2023-08-04 14:14
JMXZ
摘要:>>>vis[][] -> 减少访问 》》》dfs 要检查返回 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long
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posted @ 2023-07-29 10:49
JMXZ
摘要:只需要O(n)就可以完成 单点查询+区间修改 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define
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posted @ 2023-07-29 09:37
JMXZ
摘要:#include<bits/stdc++.h> using namespace std; int c[4],m[4]; int main() { for(int x=1;x<=3;x++)cin>>c[x]>>m[x]; for(int x=1;x<=100;x++){ int f=(x-1)%3+
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posted @ 2023-07-28 10:18
JMXZ
摘要:#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bits/stdc++.h> #define ll long long #define ddd printf(" debug\n");
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posted @ 2023-07-24 10:09
JMXZ