zhousilijames的专栏

#include #include#include#include#includeusing namespace std;int n,t,num;char s[300],c[30]="2223334445556667777888999";int main(){ while(~scanf("%d",&n)) { priority_queue, gr

#include #include#includeusing namespace std;int n,a[1000005];int readin(){ char c; int ans=0; while(c=getchar()) { if(c>='0'&&c<='9') { ans=ans*10+c

#include #include#includeusing namespace std;int m[5][6],ans[5][6],n,cnt,f[6],c[5][6];int dx[5]={0,0,1,-1,0},dy[5]={1,-1,0,0,0};void res(int x,int y){ if(c[x][y]) c[x][y]=0; els

#include #include#include#include#includeusing namespace std;struct P{ int x,y; P(){} P(int x,int y):x(x),y(y){} P operator - (const P &t) { return P(x-t.x,y-t.y);

刚开始我还很怀疑我的想法，后来百度没发现题解，看到discuss里面说数据很弱，于是我就大胆的写了个暴力,每次枚举一个点找到一次到合法终点的合法路径是，结束dfs，简单暴力代码如下#include #include#includeusing namespace std;const int mx=111;int mp[mx][mx],T,n,m,num,ans;bool use

DescriptionIn order to understand early civilizations, archaeologists often study texts written in ancient languages. One such language, used in Egypt more than 3000 years ago, is based on charact

#include #include#include#includeusing namespace std;int n,L,R;int main(){ scanf("%d",&n); scanf("%d%d",&L,&R); int x=1; while(x<R) { x*=2; } x=x/n+1;