翻过这座山，他们就会听到你的故事

链接：https://www.nowcoder.com/acm/contest/199/A来源：牛客网将认识关系转化为图中的边。dfs这张图，对每一个没有被访问过的点，将它标记为源点的反色，回溯的时候统计每个点有多少同色相邻点，个数等于2时将其颜色转换。#include<cstdio>#include<cstring>const int N=1e5+10,M...

题目链接#include<cstdio>#include<queue>#include<cstring>using namespace std;const int N=1e6+10,mod=998244353;struct node{ int u,d,pre,rnk; node(){} node(int _u,int _d,i...

题目链接#include<cstdio>#define re registertypedef long long ll;const int N=1e6+10;const int INF=0x3f3f3f3f;const int mod=1e9+7;template<typename tp>inline int getmin(tp&x,tp...

题目链接看了题解后还是没写对，只能去看Komachi大佬咋写的了。#include<cstdio>#include<cstring>const int N=18,MX=(1<<18)+5;int n,m,k,ban[N][MX],pre[MX],f[MX];int main(){ //freopen("in.txt","r",std...

题目链接#include<cstdio>#include<algorithm>#include<cmath>using namespace std;typedef long long ll;#define re registerconst int mod=1e9+7;const int N=210;template<typenam...

题目链接#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;inline int read(){ int s=0,f=0;char ch=getchar(); while(ch<'0'||ch&g...

#include<cstdio>#include<algorithm>using namespace std;const int N=2e3+10;const int mod=998244353;int n,p[N][4],w[4][4],pre[N][N],suf[N][N],ans[N];int main(){ //freopen(...

题目链接#include<cstdio>#include<algorithm>using namespace std;const int N=210;int n,m,x,y,ans,t,nk,k,a[N][N],b[N][N],vis[N][N];int main(){ //freopen("in.txt",

题目链接#include<cstdio>const int N=5e5+10;inline int read(){ int s=0,f=0;char ch=getchar(); while(ch<'0'||ch>'9')f|=ch=='-',ch=getchar(); while(ch>='0'&&ch<='9')s=(s...

题目链接#include<cstdio>#include<bitset>using namespace std;const int N=110;int n,m,q;bitset<N>f[N][31];int main(){ //freopen("in.txt","r",stdin); scanf("%d%d",&n,&

题目链接#include<cstdio>typedef long long ll;int t,k,l,r;ll len,f[4000010];int main(){ //freopen("in.txt","r",stdin); scanf("%d",&t);f[0]=1;f[1]=2; while(t--) { scanf("%d%lld",&

题目链接#include<cstdio>typedef long long ll;int t,find;ll l,r,k,x;int main(){ //freopen("in.txt","r",stdin); scanf("%d",&t); while(t--) { find=0; scanf("%lld%ll...

题目链接#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const int MAXN=1e7+10;ll a[MAXN];int r[MAXN],N,ans=1;#define RG register#define getc() (S == T ...

题目链接#include<cstdio>#include<algorithm>#include<queue>using namespace std;const int N=3e3+10;struct node{ int p,l,r; bool operator <(const node&rhs)const{ retur...

题目链接#include<cstdio>#include<algorithm>using namespace std;const int N=1e6+10,Q=1e4+10,INF=0x3f3f3f3f;int n,m,a[N],fa[21],l,r,f1,f2,ans,tot,flag,q,askl[Q],askr[Q],minl=INF,maxr=-...

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;template<typename tp>inline void read(tp&x){ x=0;int f=0;char ch=getchar(); while(ch<'...

题目链接#include<cstdio>#include<algorithm>using namespace std;int main(){ long long n,m; scanf("%lld%lld",&n,&m); if(n<m)swap(n,m); if(m<=2)printf("%lld\n",m*n);...

链接：https://www.nowcoder.com/acm/contest/199/B来源：牛客网学习了大佬代码。对着这份代码看了一个多小时好像有点点明白。大概就是在这头选两个端点在那头选两个端点更新答案（说了等于没说）。选子树内那一头是不能在同一个子节点的子树内部选两个（会多占一些边），选子树外那头也得把共了边的部分减去，保证最后往上跳之后求出来的路径组合长度是在范围内的。#in...

链接：https://www.nowcoder.com/acm/contest/199/C来源：牛客网我们枚举不同数字的个数 xxx 。此时等价于这个问题，有 x 个箱子排成一排，任意两个箱子之间距离不超过 k（超过 k 意味着可以把这个间距减小到 k，且是一个等价的序列），第一个箱子和最后一个箱子的距离不超过 m 的方案数。设 F[i,j]F[i,j]F[i,j] 表示放置了 ii...

题目链接构造一个序列 BBB 。每次二分一个答案 ansansans，构造 BBB 序列方法：Bi={−1Ai<ans1Ai≥ansB_i=\begin{cases}-1\quad A_i<ans\\1\quad A_i\geq ans\end{cases}Bi​={−1Ai​<ans1Ai​≥ans​然后维护 BBB 的前缀和，并记录前缀和的最小值。通...

题目链接#include<cstdio>#include<cstring>typedef long long ll;int n,m,t,ans[5]={0,1,5,11,36};struct Matrix{ ll a[5][5];};Matrix operator *(Matrix a,Matrix b){ Matrix c;memset(c.a...

题目链接#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N=2e3+10;int n,dp[N][N],w[N],b[N],c1[N][N],c2[N][N];int main...

题目链接学习了大佬代码，毕竟本蒟蒻至今写不来二分图匹配（NOIP退役预定qwq）#include<cstdio>#include<cstring>const int N=500;int t,n,m,g[N][N],hd[N],tot,To[N],match[N],vis[N],lim;struct Edge{ int v,nx;}e[N*N];...

题目链接#include<cstdio>#include<cstring>typedef long long ll;#define _rep(i,a,b) for(int i=(a);i<=(b);i++)#define _for(i,a,b) for(int i=(a);i<(b);i++)int co[10][10];ll pw[10]...

题目链接___#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=2e5+10;int n,m,hd[N],tot,dep[N],fa[N][20],ord[N],cnt,q;struct Edge{ int v,nx;...

题目链接我们把方差公式进行化简。记 sum1sum_1sum1​ 为每个数的前缀和，sum2sum_2sum2​ 为每个数平方后的前缀和1m∑i=1m(bi−b‾)2=1m∑i=1m(bi2−2∗bi∗b‾+b‾2)=1m(sum2−2∗b‾∗sum1+m∗b‾2)=1m(sum2−2∗sum12m+sum12m)=sum2m−sum12m2\frac{1}{m}\sum_{i=1}^m...

#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const ll mod=998244353;ll m,l1,l2,r1,r2;ll get(ll l,ll r){return r/m-(l==0?-1:(l-1)/m);}ll cal(ll l,l...

题目链接#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const int N=5e5+10,mod=998244353;inline int read(){ int s=0,f=0;char ch=getchar(); while(ch<...

题目链接#include<cstdio>const int N=4e3+10,mod=1e9+7;int a,b,c,d,fac[N],inv[N],ans;int C(int n,int m){ if(n<0||m<0)return 0; return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}int main()...

题目链接#include<cstdio>const int N=1e3+10;char s[N][N];int n,m,t;int main(){ //freopen("in.txt","r",stdin); scanf("%d",&t); while(t--) { scanf("%d%d",&n,&amp

题目链接#include<cstdio>typedef long long ll;const int mod=1e9+7;int n;ll ans=1,a,b,c;int main(){ scanf("%d",&n); if(n==1){puts("0");return 0;} a=1;b=n-2; for(int i=...