Problem A Thanks, TuSimple!
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5969
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4090
题解：二分+思维
1、配对的人喜好肯定不一样；
2、按不同喜好对身高排序；
3、O(n)方法匹配
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=300000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
char str;
struct node{
int h,flag;
bool operator <(const node&S)const{
if(flag==S.flag){
return h<S.h;
}
return flag<S.flag;
}
}a[N],b[N];
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%d",&a[i].h);
}
for(int i=1;i<=m;i++){
scanf("%d",&b[i].h);
}
for(int i=1;i<=n;i++){
scanf("%d",&a[i].flag);
}
for(int i=1;i<=m;i++){
scanf("%d",&b[i].flag);
}
sort(a+1,a+n+1);
sort(b+1,b+m+1);
ans=0;
l=n+1;
for(int i=1;i<=n;i++){
if(a[i].flag){
l=i;
break;
}
}
for(int i=1;i<=m;i++){
if(l>n)
break;
if(!b[i].flag){
if(a[l].h<b[i].h){
ans++;
l++;
}
}else{
break;
}
}
l=m+1;
for(int i=1;i<=m;i++){
if(b[i].flag){
l=i;
break;
}
}
for(int i=1;i<=n;i++){
if(l>m)
break;
if(!a[i].flag){
if(b[l].h<a[i].h){
ans++;
l++;
}
}else{
break;
}
}
cout<<ans<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem B Potion
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5970
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4091
C++版本一
题解：高精度
1、对原数循环除以2；
2、添加每次的结果到ans数组；
3、数据范围二进制大概2的3400次；
4、代码不是最优（比如如果原数已经0，可以直接退出）；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N],c[N];
char str[N];
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
//scanf("%d",&n);
cin>>str;
int len=strlen(str);
memset(c,0,sizeof(c));
memset(a,0,sizeof(a));
for(int i=1;i<=len;i++){
a[i]=str[len-i]-'0';
}
c[0]=1;
for(int i=0;i<3400+10;i++){
int now=0;
for(int j=len;j>=1;j--){
a[j]+=now*10;
now=0;
if(a[j]%2==1){
now=1;
}
a[j]/=2;
}
int limit=max(len,c[0])+100;
for(int j=1;j<=limit;j++){
c[j]+=a[j];
if(c[j]>=10){
c[j+1]+=c[j]/10;
c[j]%=10;
}
if(c[j]){
c[0]=max(c[0],j);
}
}
}
for(int i=c[0];i>=1;i--){
cout<<c[i];
}
cout<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Python版本一
t = int(input())
while t > 0:
n = int(input())
ans = int(0)
t -= 1
while n > 0:
n //= 2
ans += n
print(ans) Problem C Robot Cleaner I
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5971
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4092
C++版本一
题解：
1、有个东西叫%1d；
2、哈希坐标；
3、剪枝；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
//#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=2000+100;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m;
ll k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a,b,mp[N][N];
int vis[N][300];
char str[N],str1[N];
struct node{
int x,y,step;
}tmp,start,front1;
int command(int x,int y){
return 81*mp[x][y]+27*mp[x-1][y]+9*mp[x+1][y]+3*mp[x][y-1]+mp[x][y+1];
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
scanf("%d%d%lld",&a,&b,&k);
scanf("%s",str);
sum=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%1d",&mp[i][j]);
if(mp[i][j]==2)
sum++;
}
}
start.x=a;
start.y=b;
start.step=0;
memset(vis,-1,sizeof(vis));
ans=0;
front1=start;
while(1){
int opid=command(front1.x,front1.y);
char op=str[opid];
if(vis[(front1.x-1)*m+front1.y][opid]==ans)
break;
if(sum<=ans)
break;
vis[(front1.x-1)*m+front1.y][opid]=ans;
tmp=front1;
tmp.step++;
if(tmp.step>k)
break;
//cout<<front1.x<<" "<<front1.y<<" "<<op<<" "<<mp[front1.x][front1.y]<<" "<<vis[front1.x][front1.y]<<endl;
if(op=='U'){
tmp.x--;
}else if(op=='D'){
tmp.x++;
}else if(op=='L'){
tmp.y--;
}else if(op=='R'){
tmp.y++;
}else if(op=='P'){
if(mp[front1.x][front1.y]==2){
mp[front1.x][front1.y]=0;
ans++;
}
}else if(op=='I'){
break;
}
if(mp[tmp.x][tmp.y]==1){
break;
}
front1=tmp;
}
printf("%d\n",ans);
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
题解：注意标记，vis[i][j] i表示对坐标的哈希，j表示当前计算的x，值为之前得到多少汽油了，因为如果再次经过这个位置，两次经过的中间得到多汽油，并且改变了图中的2，所以要继续走，知道，走到一个位置，汽油数不增加，然后就是注意一些剪枝。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m;
int a,b;
ll k;
char mp[2010][2010];
int vis[2100][255];
char step[255];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
scanf("%d%d%lld",&a,&b,&k);
scanf("%s",step+1);
int sum=0;
for(int i=0;i<=n*m;i++)
for(int j=0;j<=250;j++)
vis[i][j]=-1;
for(int i=1;i<=n;i++)
{
scanf("%s",mp[i]+1);
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='2')
sum++;
}
}
int ans=0;
int x;
int xx,yy;
int st=0;
while(1)
{
x=81*(mp[a][b]-'0')+27*(mp[a-1][b]-'0')+9*(mp[a+1][b]-'0')+3*(mp[a][b-1]-'0')+(mp[a][b+1]-'0');
if(vis[(a-1)*m+b][x]==ans) break;
vis[(a-1)*m+b][x]=ans;
x++;
// cout<<a<<" "<<b<<" "<<x<<" "<<step[x]<<" "<<endl;
if(step[x]=='U') xx=a-1,yy=b;
if(step[x]=='D') xx=a+1,yy=b;
if(step[x]=='L') xx=a,yy=b-1;
if(step[x]=='R') xx=a,yy=b+1;
if(step[x]=='P')
{
if(mp[a][b]=='2')
{
ans++;
mp[a][b]='0';
}
if(ans>=sum) break;
xx=a,yy=b;
}
if(step[x]=='I')
{
xx=a,yy=b;
break;
}
if(mp[xx][yy]=='1') break;
st++;
if(st==k) break;
a=xx;
b=yy;
}
printf("%d\n",ans);
}
return 0;
}
Problem E Potion
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5973
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4094
题解：
1、向上依次寻找可代替药水；
2、如果当前不能满足则找不到；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N],b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(int i=1;i<=n;i++){
scanf("%d",&b[i]);
}
ans=1;
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
if(a[i]>b[j]){
a[i]=a[i]-b[j];
b[j]=0;
}else{
b[j]=b[j]-a[i];
a[i]=0;
break;
}
if(j==n){
ans=0;
}
}//cout<<a[i]<<endl;
}
cout<<(ans?"Yes":"No")<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem G Postman
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5975
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4096
题意：x=0的位置是邮局，邮递员每次可以带k封信，需要回邮局拿信，最后送完信后不必回邮局，求最小路程
C++版本一
题解：
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define FI first
#define SE second
#define PB push_back
typedef pair<int,LL>PII;
const int N=2e5+7;
const LL INF=1e17,mod=998244353;
int n,m,k;
LL a[N],b[N];
LL ans;
int main()
{
int t;
cin>>t;
while(t--){
ans=0;
scanf("%d%d",&n,&k);
LL x;
int cnt=0,cot=0;
for(int i=1;i<=n;i++){
scanf("%lld",&x);
if(x==0)continue;
if(x<0)a[++cnt]=-x;
else b[++cot]=x;
}
sort(a+1,a+cnt+1);
sort(b+1,b+cot+1);
if(a[cnt]>b[cot]){
ans+=a[cnt];
cnt=max(cnt-k,0);
}
else {
ans+=b[cot];
cot=max(cot-k,0);
}
//cout<<ans<<endl;
while(cnt>0||cot>0){
if(a[cnt]>b[cot]){
ans+=a[cnt]*2;
cnt=max(cnt-k,0);
}
else{
ans+=b[cot]*2;
cot=max(cot-k,0);
}
}
printf("%lld\n",ans);
}
return 0;
}
C++版本二
题解：
1、从两边往中间走；
2、减掉一个绝对值最大的点；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
ll c[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
srand((unsigned)time(NULL));
scanf("%d",&t);
//t=rand()%(2000)+10000;
while(t--){
scanf("%d%d",&n,&k);
ll maxl=0;
for(int i=1;i<=n;i++){
scanf("%lld",&c[i]);
maxl=max(maxl,abs(c[i]));
}
sort(c+1,c+n+1);
int l=0;
int r=n+1;
ans=0;
for(int i=n;i>=1;i--){
if(c[i]>0){
r=i;
}else{
break;
}
}
for(int i=1;i<=n;i++){
if(c[i]<0){
l=i;
}else{
break;
}
}
for(int i=n;i>=r;i-=k){
ans+=2ll*abs(c[i]);
}
for(int i=1;i<=l;i+=k){
ans+=2ll*abs(c[i]);
}
cout<<ans-maxl<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem J Extended Twin Composite Number
比赛地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5977
补题地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4099
题意：找到两个数x，y使得满足x+n=y，并且x，y都是合数
题解：构造
1、特判n==1时；
2、任何数*2肯定合数，+n就是3*n，也肯定合数；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%lld",&t);
while(t--){
scanf("%lld",&n);
if(n==1){
cout<<9<<" "<<10<<endl;
continue;
}
cout<<2*n<<" "<<3*n<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
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