bettle_king的博客

https://www.acwing.com/problem/content/description/1127/题目问的是，新旧牧场的最大直径中的最小直径。需要注意的是，牧场有很多个，我们只连接一条边，故最多是两个连通块合并。maxv[i] 表示一个联通块中距离i最远的距离挨个枚举所有的不再一个连通块中的点。#include<bits/stdc++.h>using namespace std;const int N=200;const double INF=1e12;dou.

http://acm.hdu.edu.cn/showproblem.php?pid=1205找到最多的球。俩俩不同的配对。#include<bits/stdc++.h>using namespace std;typedef long long int LL;int main(void){ int t; cin>>t; while(t--) { LL n,x; scanf("%lld",&n); LL sum=0,maxv=0; for(in.

http://acm.hdu.edu.cn/showproblem.php?pid=1196#include<cstdio> #include<iostream>using namespace std;int lowerbit(int x){ return x&(-x);}int main(void){ int n; while(cin>>n,n!=0) { cout<<lowerbit(n)<<endl; .

https://codeforces.com/problemset/problem/1567/B根据MEX的定义可以知道的是 [0,a-1]必定全部存在，a一定不在。我们可以先预处理出所有的前缀异或和。#include<bits/stdc++.h>using namespace std;const int N=1e5*3+10;int s[N];void init(){for(int i=1;i<N;i++) s[i]=s[i-1]^i;}//预处理int main(vo.

https://www.acwing.com/problem/content/1295/其实你会发现，质数和其合数必有一条边。而质数之间无联系，合数之间也无联系。这就是一个二分图问题。那么如果有合数，则结果必为二，将合数染成同一个颜色，质数染成同一个颜色。否则只有质数，则都是同一个颜色。#include<bits/stdc++.h>using namespace std;const int N=1e5+10;int n,a[N],flag;bool check(int x).

https://codeforces.com/problemset/problem/270/A所以n边形的内角和是（n-2）·180°.（n为边数）先打表符合的所有的角度#include<bits/stdc++.h>using namespace std;map<double,int>mp;int main(void){ for(int i=3;i<=700;i++) mp[1.0*(i-2)*180/i]++; int t; cin>>t; .

https://codeforces.com/problemset/problem/441/A#include<bits/stdc++.h>using namespace std;int n,m;vector<int>ans;int main(void){ cin>>n>>m; for(int i=0;i<n;i++) { int k; cin>>k; bool flag=0; for(int j=0;j&lt.

https://leetcode-cn.com/problems/construct-the-rectangle/class Solution {public: vector<int> constructRectangle(int area) { int L=area,W=1; for(int i=1;i<=area/i;i++) if(area%i==0) L=area/i,W=i; vector<int>ans;.

题目地址#include<cstdio>#include<iostream>using namespace std;int a[252],cnt[252];int main(void){ int n,m; cin>>n>>m; for(int i=0;i<n;i++) cin>>a[i],cnt[a[i]]++; for(int i=0;i<n;i++) if(cnt[a[i]]>=2.

https://www.nowcoder.com/practice/0383714a1bb749499050d2e0610418b1?tpId=40&tqId=21333&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<algorithm>.

https://www.nowcoder.com/practice/fe8bff0750c8448081759f3ee0d86bb4?tpId=40&tqId=21358&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<iostream>#include<algorithm>#include<cstdio>.

https://www.nowcoder.com/practice/7557d470617c464f9200d93acf721471?tpId=40&tqId=21462&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<string>#i..

https://www.nowcoder.com/practice/139761e0b59a405786898b7f2db9423f?tpId=40&tqId=21425&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<string>#in.

https://www.nowcoder.com/practice/a9fa482eb05149cdbad88689e3cb5f66?tpId=40&tqId=21551&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<string>#in.

https://www.nowcoder.com/practice/bbbbf26601b6402c9abfa88de5833163?tpId=40&tqId=21398&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<algorithm>.

https://www.nowcoder.com/practice/3f27a0a5a59643a8abf0140b9a8cf1f7?tpId=40&tqId=21340&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<algorithm>.

https://www.nowcoder.com/practice/27fbaa6c7b2e419bbf4de8ba60cf372b?tpId=40&tqId=21368&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking&tab=answerKey#include<cstdio>#include<iostream>#include<algorithm>.

http://acm.hdu.edu.cn/showproblem.php?pid=2007没有规定 x一定小于 y#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;int main(void){ int a,b;.

http://acm.hdu.edu.cn/showproblem.php?pid=2629地区只要看前两位就可以了，因为地区是个大的概念后面的数字代表的是县#include<cstdio>#include<iostream>#include<string>#include<algorithm>using namespace std;int main(void){ int n; cin>>n; while(n--) { .

http://acm.hdu.edu.cn/showproblem.php?pid=1012#include<cstdio>#include<iostream>using namespace std;int main(void){ cout<<"n e"<<endl; cout<<"- -----------"<<endl; for(int i=0;i<=9;i++) { .

http://acm.hdu.edu.cn/showproblem.php?pid=1197#include<cstdio>#include<iostream>using namespace std;int check1(int x){ int sum=0; while(x) { sum+=x%10; x/=10; } return sum;}int check2(int x){ int s[100]; int ans=0; int sum=.

http://acm.hdu.edu.cn/showproblem.php?pid=2018#include<cstdio>#include<iostream>using namespace std;const int N=60;long long int s[N];int main(void){ int n; s[1]=1,s[2]=2,s[3]=3; while(cin>>n,n!=0) { for(i.

http://oj.ecustacm.cn/problem.php?id=1098#include<cstdio>#include<iostream>#include<set>#include<map>#include<algorithm>using namespace std;set<int> st;map<int,int> mp;int main(void){ int n; cin>>.

https://www.luogu.com.cn/problem/P1075看题目是两个质数的乘积如果从 n-1 枚举到 2会超时。 因为是成对出现的所以找到小的 n/小的 就找到了#include<cstring>#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<set>#include<cmat.

https://www.luogu.com.cn/problem/P1059#include<cstring>#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<set>using namespace std;int a[101];set<int>st;int main(void){ int n.

https://www.luogu.com.cn/problem/P1046#include<cstring>#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[11];int main(void){ int i=0; int max; for(int i=0;i<10;i++).

P1035 [NOIP2002 普及组] 级数求和https://www.luogu.com.cn/problem/P1035#include<cstdio>int main(void){ int k; scanf("%d",&k); double sum=1; long int j=2; while(sum<=k)//注意小于等于 { sum+=1.0/j; j++; } printf("%ld\n",j); return 0;}...