J - Anti-merge 黑白染色
从第一个位置开始遍历，他四周如果有和他相同的就把他染成和自己不一样的颜色，之后加进队列知道最后遇不到相同的数；这样从第一个位置开始就形成了一个连通块，看一看用哪一个颜色比较少就用哪一个，因为如果一个位置的颜色与四周的不一样，四周的是不能和这个位置合并的，四周的颜色却可以相等，所以只需要一种标签就可以了；
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
ll n,m,a[510][510],vis[510][510];
ll dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
struct node{
ll x,y,col;
};
vector<pair<ll,ll>>ans;
void bfs(ll x,ll y){
queue<node>q;
q.push(node{x,y,0});
vector<pair<ll,ll>>w,b;
//w装的是标签为1的，b装0的
vis[x][y]=1;
b.push_back({x,y});
while(!q.empty()){
node u=q.front();q.pop();
for(int i=0;i<4;i++){
ll tx=u.x+dx[i],ty=u.y+dy[i];
if(tx<=n&&tx>=1&&ty<=m&&ty>=1&&a[u.x][u.y]==a[tx][ty]&&!vis[tx][ty]){
vis[tx][ty]=1;
// cout<<tx<<" "<<ty<<" "<<u.x<<" "<<u.y<<endl;
if(u.col==1) b.push_back({tx,ty});
else w.push_back({tx,ty});
q.push(node{tx,ty,!u.col});
}
}
// cout<<"wb "<<w.size()<<" "<<b.size()<<endl;
}
if(b.size()>=w.size()){
for(auto x:w) ans.push_back(x);
}
else {
for(auto x:b) ans.push_back(x);
}
}
int main(){
// freopen("in.txt","r",stdin);
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) scanf("%lld",&a[i][j]),vis[i][j]=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) if(!vis[i][j]) bfs(i,j);
ll type=(ans.size()==0)?0LL:1LL;
printf("%lld %d\n",type,ans.size());
for(auto x:ans){
printf("%lld %lld %lld\n",x.first,x.second,type);
}
return 0;
}
/*
10 10
7 3 6 8 4 5 1 2 8 7
2 5 7 6 1 8 8 6 7 6
7 1 5 6 6 4 5 7 3 3
8 8 8 6 1 7 7 3 3 1
4 7 7 2 8 7 8 4 3 4
3 5 2 6 8 4 1 8 6 5
2 5 7 6 7 8 7 8 4 2
3 2 1 8 7 2 7 3 1 6
8 5 4 2 3 8 7 2 8 1
8 3 2 5 3 5 3 2 7 3
*/
整除光棍 - 题目 - Daimayuan Online Judge
不用找规律，也不需要打表，只需要模拟除法即可，逐渐增加位数，看看是否能够整除x
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
ll x,ans[100005],cnt;
string fac[10005];
ll weis(ll y){
ll res=0;
while(y){
res++;
y/=10;
}
return res;
}
bool check(string s){
for(int i=0;i<=cnt;i++) ans[i]=0;
cnt=0;ll shang=0,yu=0,chu=0,pos=0,cut=0;
while(pos<s.size()){
ll ct=0;
while(chu<x&&pos<s.size()){
chu=chu*10+s[pos]-'0';
pos++;
if(ct>=1&&cut>=1) ans[++cnt]=0;
ct++;
}
cut++;
shang=chu/x; ans[++cnt]=shang;
yu=chu%x;
//cout<<shang<<" "<<chu<<" "<<yu<<endl;
chu=yu;
if(yu==0) break;
}
if(yu==0) return 1;
else return 0;
}
int main(){
// freopen("in.txt","r",stdin);
fac[1]="1";
for(int i=2;i<=10005;i++) fac[i]=fac[i-1]+"1";
scanf("%lld",&x);
ll tmp=x;
tmp=weis(tmp); cnt=0;
for(int i=tmp;i<=10000;i++){
//cout<<i<<" "<<fac[i]<<endl;
if(check(fac[i])){
for(int i=1;i<=cnt;i++)
cout<<ans[i];
cout<<" ";
cout<<i<<endl;
break;
}
}
return 0;
}
D - Pattern Lock
以此题来记录一下自己贫弱的代码能力，，，
先考虑n是偶数的情况，不是偶数就留下后面三行，然后在考虑m是偶数的情况，不是就留下后三列；
(7条消息) 2021 Jiangsu Collegiate Programming Contest 总结_Evil_boy__的博客-CSDN博客
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
ll n,m;
int main(){
// freopen("in.txt","r",stdin);
scanf("%lld%lld",&n,&m);
if(n%2==0){
for(ll i=1;i<=n;i+=2){
printf("%d 1\n",i);
for(ll j=1;j<m;j++){
printf("%lld %lld\n",i,j+1);
printf("%lld %lld\n",i+1,j);
}
printf("%lld %lld\n",i+1,m);
}
}
else{
for(ll i=1;i<=n-3;i+=2){
printf("%lld 1\n",i);
for(ll j=1;j<m;j++){
printf("%lld %lld\n",i,j+1);
printf("%lld %lld\n",i+1,j);
}
printf("%lld %lld\n",i+1,m);
}
if(m%2==0){
for(ll i=m-1;i>=1;i-=2){
printf("%lld %lld\n",n-2,i);
printf("%lld %lld\n",n-2,i+1);
printf("%lld %lld\n",n-1,i);
printf("%lld %lld\n",n-1,i+1);
printf("%lld %lld\n",n,i);
printf("%lld %lld\n",n,i+1);
}
}
else{
for(ll i=m-1;i>=3;i-=2){
printf("%lld %lld\n",n-2,i);
printf("%lld %lld\n",n-2,i+1);
printf("%lld %lld\n",n-1,i);
printf("%lld %lld\n",n-1,i+1);
printf("%lld %lld\n",n,i);
printf("%lld %lld\n",n,i+1);
}
cout << n - 1 << " " << 2 <<endl;
cout << n - 2 << " " << 3 << endl;
cout << n << " " << 2 << endl;
cout << n - 2 << " " << 1 << endl;
cout << n - 1 << " " << 3 << endl;
cout << n << " " << 1 << endl;
cout << n - 2 << " " << 2 << endl;
cout << n << " " << 3 << endl;
cout << n - 1 << " " << 1 << endl;
}
}
return 0;
}
Binary String To Subsequences
vector存储第cnt个字串最后一个元素的值，a[i]记录第i个元素在第几个vector中。拿一个one记录最后一个1的vector的位置，zero记录第一个0的位置，如果是0且one>0，说明还存在至少one个vector末尾的元素是1，则a[i]=one,one--，若one<=0,说明现在是没有了，要新开一个vector了，a[i]=++cnt；若是1且zero<=cnt,则说明还有vector末尾是0，则让a[i]=zero,zero++,one--，因为存0的vector变少了，存1的变多了嘛；最后输出cnt和a[i]即可
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
ll n,t,a[200005];
string s;
vector<ll>v[200005];
int main(){
// freopen("in.txt","r",stdin);
scanf("%lld",&t);
while(t--){
scanf("%lld",&n);
for(int i=1;i<=n;i++) a[i]=0,v[i].clear();
ll cnt=0,one=0,zero=2;
cin>>s;s=" "+s;
v[++cnt].push_back(s[1]-'0'); a[1]=cnt;
if(s[1]=='1') one++;else zero=1;
for(int i=2;i<=n;i++){
if(s[i]-'0'==0){
if(one>0){
v[cnt].push_back(s[i]-'0');
a[i]=one;
one=max(0LL,one-1);
zero=max(1LL,zero-1);
}
else {
v[++cnt].push_back(s[i]-'0');
a[i]=cnt;
}
}
else{
if(zero>=1&&zero<=cnt){
v[zero].push_back(s[i]-'0');
a[i]=zero;
zero++;one++;
}
else{
v[++cnt].push_back(s[i]-'0');
a[i]=cnt;
zero++;one++;
}
}
}
printf("%lld\n",cnt);
for(int i=1;i<=n;i++) printf("%lld ",a[i]);
printf("\n");
}
return 0;
}
