飞机票：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=505

直接进行BFS就行，需要注意的是状态转移，对四个点进行排序，然后按100，10000，1000000的系数来判断是否访问过(剪枝)，因为一个点可以上下左右跳一格或两格如果一格可以则不进行两格(剪枝)//

//  main.cpp
//  zoj1505
//
//  Created by cfhaiteeh on 12/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
struct he{
    int ai[9];
    int val;
    int t;
};
int bi[9];
int ci[9];
map<int,int> _M;
int _map[9][9];
int rn[9];
int cn[9];
int toans;
bool getdir(int r,int c){
    if(r<1||r>8)return 0;
    if(c<1||c>8)return 0;
    if(_map[r][c]==1)return 0;
    return 1;
}
int cm[5];
int cr[]={-1,-2,1,2,0,0,0,0};
int cc[]={0,0,0,0,-1,-2,1,2};
int getcm(){
    sort(cm+1,cm+1+4);
    return cm[1]*1000000+cm[2]*10000+cm[3]*100+cm[4];;
}
bool BFS(){
    queue<he> Q;
    he ne;
    for(int i=1;i<=8;i++)ne.ai[i]=bi[i];
    for(int k=2;k<=8;k+=2){
        cm[k/2]=ne.ai[k-1]*10+ne.ai[k];
    }
    int ct=getcm();
    _M[ct]=1;
    ne.val=8;
    ne.t=ct;
    Q.push(ne);
   
    while(!Q.empty()){
        he c=Q.front();
        Q.pop();
        if(c.t==toans)return 1;
        if(c.val==0)continue;
    
        memset(_map,0,sizeof(_map));
        for(int i=1;i<=8;i+=2){
            int row=c.ai[i];
            int col=c.ai[i+1];
            _map[row][col]=1;
           
        }
        for(int i=1;i<=8;i+=2){
            int row=c.ai[i];
            int col=c.ai[i+1];
    
            int st=0;
            for(int j=0;j<8;j++){
                if(getdir(row+cr[j],col+cc[j])){
                    rn[++st]=row+cr[j];
                    cn[st]=col+cc[j];
                    if(j%2==0)
                    {
                        j++;
                    }
                }
            }
       
            for(int j=1;j<=st;j++){
                
                    he nex;
                    for(int k=1;k<=8;k++){
                        nex.ai[k]=c.ai[k];
                        
                    }
                    nex.ai[i]=rn[j];
                    nex.ai[i+1]=cn[j];
                    nex.val=c.val-1;
                for(int k=2;k<=8;k+=2){
                    cm[k/2]=nex.ai[k-1]*10+nex.ai[k];
                }
             
                int ct=getcm();
                if(_M[ct]==1)continue;
                nex.t=ct;
                _M[ct]=1;
                Q.push(nex);
            }
        }
       
    }
    return 0;
}
void init(){
    _M.clear();
    for(int i=2;i<=8;i++)scanf("%d",&bi[i]);
    for(int i=1;i<=8;i++)scanf("%d",&ci[i]);
    
    for(int k=2;k<=8;k+=2){
        cm[k/2]=ci[k-1]*10+ci[k];
    }
   
    int ct=getcm();
    toans=ct;
 
}
int ans;
void play(){
    ans=BFS();
}
void pri(){
    if(ans)printf("YES\n");
    else printf("NO\n");
}
int main() {
    // insert code here...
    while(scanf("%d",&bi[1])!=EOF){
        init();
        play();
        pri();
    }
    return 0;
}

飞机票:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1347

易知对于3^i次方 3^0+3^1+....3^(i-1)<3^i 且对于k项，一共有2^k-1(不包括空集，对空集进行特殊处理) 所以当n==2^k-1时，则前k项都包括，当n>2^k-1&&n<2^(k+1)-1时，包含第k+1项，将n-(2^k-1)后就是k+1项的事情了，领n1=n-(2^k-1),

n2=n1-1时，此时k+1已经处理了，则n2又变为原来的n，进行递归求解。

import java.math.BigInteger;
import java.util.*;

public class Maintf {
	public static BigInteger fac[];
	public static ArrayList<Integer> al;
	public static void DFS(BigInteger x){
		if(x.compareTo(BigInteger.ZERO)==0)return;
		for(int i=1;i<64;i++){
			if(x.compareTo(fac[i])==0){
				for(int j=i;j>=1;j--){
					al.add(j);
				}
				return;
			}
			if(x.compareTo(fac[i])>0&&x.compareTo(fac[i+1])<0){
				x=x.subtract(fac[i]).subtract(BigInteger.ONE);
				al.add(i+1);
				DFS(x);
				return;
			}
		}
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		fac=new BigInteger[123];
		fac[1]=BigInteger.ONE;
		BigInteger two=new BigInteger("2");
		for(int i=2;i<=64;i++){
			fac[i]=fac[i-1].multiply(two).add(BigInteger.ONE);
		
		}
		BigInteger pow3[]=new BigInteger[123];
		pow3[1]=BigInteger.ONE;
		BigInteger three=new BigInteger("3");
		for(int i=2;i<=64;i++){
			pow3[i]=pow3[i-1].multiply(three);
		}
		al=new ArrayList<>();
		Scanner sca=new Scanner(System.in);
		while(sca.hasNext()){
			
			BigInteger n=sca.nextBigInteger();
			if(n.compareTo(BigInteger.ZERO)==0)break;
			if(n.compareTo(BigInteger.ONE)==0){
				System.out.println("{ }");
				continue;
			}
			al.clear();
			n=n.subtract(BigInteger.ONE);
			
			DFS(n);
			
			System.out.print("{");
			for(int i=al.size()-1;i>=0;i--){
				System.out.print(" "+pow3[al.get(i)]);
				if(i!=0)System.out.print(",");
				else System.out.print(" ");
			}
			System.out.println("}");
		}
	}

}

A - Diagonal String

水题：记录直接输出就好

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char ch[4][4];
int main() {
    for(int i=1;i<=3;i++)scanf("%s",ch[i]+1);
    for(int i=1;i<=3;i++)printf("%c",ch[i][i]);
    printf("\n");
    return 0;
}

B - Palindromic Numbers

数据较小，判断记录总的多少就好

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char ch[4][4];
int ai[123456];
  int ci[6];
int main() {
    for(int i=10000;i<=99999;i++){
      
        int st=0;
        int t=i;
        while(t>0){
            int p=t%10;
            t=t/10;
            ci[++st]=p;
        }
        bool flag=1;
        for(int j=1;j<=2;j++){
            if(ci[j]==ci[5-j+1]){
                continue;
            }
            flag=0;
            break;
        }
        if(flag)ai[i]=1;
    }
    for(int i=10000;i<=99999;i++)ai[i]+=ai[i-1];
    int l,r;
    while(scanf("%d%d",&l,&r)!=EOF){
        printf("%d\n",ai[r]-ai[l-1]);
    }
    return 0;
}

C - Flip,Flip, and Flip......

找规律，特判就行

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
LL n,m;
LL ans;
int main() {
    while(scanf("%lld%lld",&n,&m)!=EOF){
        if(n>m)swap(n,m);
        if(n==1&&m==1)printf("1\n");
        else {
            if(n==1)printf("%lld\n",m-2);
            else {
                LL ans=(n-2)*(m-2);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}

D - Remainder Reminder

对于除数进行枚举，易知当除数为i时，1~n的余数为1，2，。。。。0，1，2.。。。0 循环。所以只要直接判断n里有多少组1.2.。。0即可，对于不能整除时进行特判就行。复杂度为O（n）本人对0也进行了特判，主要感觉0怪怪的。。。//

//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
LL n,m;
LL ans;
int main() {
    while(scanf("%lld%lld",&n,&m)!=EOF){
        ans=0;
        if(m==0){
            ans=n;
            for(int i=2;i<=n;i++){
                LL p=n/i;
                ans+=p;
            }
            m=1;
        }
       
        
            for(LL i=m+1;i<=n;i++){
                LL p=n/i;
                LL q=(n%i)-m+1;
                if(q<0)q=0;
                LL z=i-m;
                ans=ans+z*(p)+q;
                
            }
            printf("%lld\n",ans);
        
    }
    return 0;
}

飞机票：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1273

对于每个回合，求出每个回合第一个数字是多少，则第i回合增量为2^i的等差数列。然后对于位置进行处理，算出当前数字的位置中最大能到达的层数。该层数和前一个数字的层数比较，选出最大的即为答案

//
//  main.cpp
//  zoj2273
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define  LL long long
using namespace std;

int turn[5*123456];
bool vis[5*123456];
int ans[100000];
int tpow[5*123456];
int wzz[100000];
void getTurn(){
    int c=1;
    int now=1;
    int t=2;
    int before=1;
    int z=0;
    while(true){
        now=now^1;
        if(now==0){
            int st=c+before;
            for(int i=st;i<=500000;i+=t){
                vis[i]=1;
            }
            tpow[++z]=t;
            t*=2;
            turn[z]=st;
           
        }else{
            for(int i=c;i<=500000;i+=t){
                vis[i]=1;
            }
            before=t;
            tpow[++z]=t;
            t*=2;
            bool flag=1;
             turn[z]=c;
            for(int i=c;i<=500000;i++){
                if(!vis[i]){
                    c=i;
                    flag=0;
                    break;
                }
            }
           
        
            if(flag)break;
        }
        
    }
  
}
void getAw(){
    ans[1]=1;
    ans[2]=1;
    ans[3]=3;
    wzz[1]=1;
    wzz[2]=1;
    wzz[3]=3;
    int s=3;
    int ci[6];
    for(int i=4;i<=99999;i++){
        int tp=i;
        int op=0;
        int now=-1;
        int ip=ans[i-1];
        int ww=wzz[i-1];
        for(int k=1;k<=500000;k++){
            int a=wzz[i-1]-turn[k];
            if(a<0)continue;
          
            if(a%tpow[k]==0){
                now=k;
                break;
            }
        }
        while(tp>0){
            int c=tp%10;
            ci[++op]=c;
            tp=tp/10;
        }
        for(int j=op;j>=1;j--){
            s++;
            int wz=s;
          
            for(int k=1;k<=500000;k++){
                int a=wz-turn[k];
           
                if(a<0)continue;
                if(a%tpow[k]==0){
                    
                    if(k>now){
                       
                        now=k;
                        ip=ci[j];
                        ww=wz;
                    }
                    break;
                }
            }
           
        }
        ans[i]=ip;
        wzz[i]=ww;
    }
    
}
int main() {
    getTurn();
    getAw();
  //  cout<<"end"<<endl;
    int a;
    while(scanf("%d",&a)!=EOF){
        printf("%d\n",ans[a]);
    }
    return 0;
}

人生就是兜兜转转，请相信自己每次的决定

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
using namespace std;
const int maxn=23;
double ai[maxn];
struct matrix
{
    double e[maxn][maxn];
};
matrix Cal(matrix a,matrix b,int len)
{
    matrix c;
    for(int i=1; i<=len; i++)
    {
        for(int j=1; j<=len; j++)
        {
            c.e[i][j]=0;
            for(int k=1; k<=len; k++)
            {
                c.e[i][j]+=a.e[i][k]*b.e[k][j];
            }
        }
    }
    return c;

}
matrix a,s;
int n;
void exp_pow(int b)
{
    while(b)
    {
        if(b&1)s=Cal(s,a,n);
        a=Cal(a,a,n);
        b>>=1;
    }
}
void init()
{   scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            s.e[i][j]=0;
        }
    }
    for(int i=1; i<=n; i++)
    {
        scanf("%lf",&s.e[1][i]);
    }
    int k,t;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            a.e[i][j]=0;
        }
    }
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&k);
        if(k==0)
        {
            a.e[i][i]=1;
        }
        else
        {
            double fm=1.0/k;
            for(int j=1; j<=k; j++)
            {
                scanf("%d",&t);
                a.e[i][t]=fm;
            }
        }
    }
}
void play()
{
    int k;
    scanf("%d",&k);
    exp_pow(k);
    bool first=1;
    for(int i=1; i<=n; i++)
    {
        if(!first)printf(" ");
        first=0;
        printf("%.2f",s.e[1][i]);
    }
    printf("\n");
}
int T;
int main()
{

    scanf("%d",&T);
    while(T--)
    {
        init();
        play();
    }
//    cout << "Hello world!" << endl;
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define INF 1e9
#define EPS 1e-9
using namespace std;
const int maxn=1234;
char ch[maxn];
int len;
int ai[maxn];
int f;
void init()
{
    scanf("%s",ch+1);
    len=strlen(ch+1);
    f=0;
    for(int i=1; i<=len; i++)
    {
        if(ch[i]=='*')ai[i]=1;
        else ai[i]=0;
        f+=ai[i];
    }
}

void play()
{
    int f2=len-f;
    f2=f-f2;
    int ans=0;
    if(f2>=0)
    {
        ans=f2+1;
    }
    int before=ans;
    int st=len;
    for(int i=1; i<=len; i++)
    {
        if(ai[i]==1)
        {
            if(before<2)
            {
                while(ai[st]!=0)
                {
                    st--;

                }
                ai[st]=1;
                before++;
                ans++;
            }
            else
            {
                before--;
            }
        }
        else
        {
            before++;
        }
    }
    if(ai[len]!=1&&f!=0)ans++;
    printf("%d\n",ans);
}
int T;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        init();
        play();
    }
//    cout << "Hello world!" << endl;
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#define LL long long
#define INF 1e9
#define EPS 1e-9
using namespace std;

const double PI=3.1415926535898;
double CalSphere(double r,double h)
{
    return PI*(r*h*h-1.0*h*h*h/3);

}
const int maxn=123456;
struct he
{
    double ch,w,l,h,r;

} Sky[maxn*2];
double w,l,v;
int n,m;
double maxH;
void init()
{
    maxH=0;
    scanf("%lf%lf%lf%d%d",&w,&l,&v,&n,&m);

    for(int i=1; i<=n; i++)
    {

        scanf("%lf%lf%lf%lf",&Sky[i].ch,&Sky[i].w,&Sky[i].l,&Sky[i].h);
        Sky[i].ch=Sky[i].ch-Sky[i].h/2;
        maxH=max(maxH,Sky[i].ch+Sky[i].h);

    }
    for(int i=1; i<=m; i++)
    {
        scanf("%lf%lf",&Sky[n+i].ch,&Sky[n+i].r);
        Sky[n+i].ch=Sky[n+i].ch-Sky[n+i].r;
        maxH=max(maxH,Sky[n+i].ch+Sky[n+i].r*2);
    }

}
bool check(double h)
{
    double nv=0;
    for(int i=1; i<=n; i++)
    {
        double mh=Sky[i].ch+Sky[i].h;
        if(mh<=h)
        {
            nv+=Sky[i].h*Sky[i].l*Sky[i].w;
        }
        else if(Sky[i].ch>=h)
        {

        }
        else
        {
            double hh=h-Sky[i].ch;
            nv+=hh*Sky[i].l*Sky[i].w;
        }
    }
    for(int i=1; i<=m; i++)
    {
        double mh=Sky[n+i].ch+Sky[n+i].r*2;
        if(mh<=h)
        {
            nv+=CalSphere(Sky[n+i].r,Sky[n+i].r*2);
        }
        else if(Sky[n+i].ch>=h)
        {

        }
        else
        {
            double hh=h-Sky[n+i].ch;
            nv+=CalSphere(Sky[n+i].r,hh);
        }
    }
    double cm=w*l*h;
    if(cm<=nv+v)
    {
        return 1;
    }
    else
    {
        return 0;
    }

}
double  BRSearch()
{

    double l=0;
    double r=maxH+1000;
    double ans=0;
    while(l<=r)
    {
        double mid=(l+r)/2;
        bool ck=check(mid);
        if(ck)
        {
            ans=mid;
            l=mid+EPS;
        }
        else
        {
            r=mid-EPS;
        }
    }
    return ans;

}
void play()
{
    double ans=BRSearch();
    if(ans>maxH)
    {
        for(int i=1; i<=n; i++)
        {
            v+=Sky[i].w*Sky[i].l*Sky[i].h;
        }
        for(int i=1; i<=m; i++)
        {
            v+=CalSphere(Sky[n+i].r,Sky[n+i].r*2);
        }
        ans=v/(w*l);
        printf("%.8f\n",ans);
    }
    else
    {
        printf("%.8f\n",ans);
    }
}
int T;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        init();
        play();
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#define LL long long
#define INF 1e9
#define EPS 1e-9
using namespace std;
const int maxn=123456;
int ai[maxn];
int Link[12345];
int bi[12345];
int T;
int n,m,k;

bool cmp(int a,int b)
{
    return a>b;
}
void init()
{

    scanf("%d%d%d",&n,&m,&k);
    memset(bi,0,sizeof(bi));
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&ai[i]);
        if(ai[i]==0)continue;
        bi[ai[i]]++;
    }
    Link[0]=0;
    bi[0]=100000000;
    for(int i=1; i<=10000; i++)
    {
        if(bi[i]==0)Link[i]=Link[i-1];
        else Link[i]=i;
    }
    sort(ai+1,ai+1+n,cmp);
}
vector<int> V;
int Find(int x)
{
    if(Link[x]==x)return x;
    Link[x]=Find(Link[x]);
    return Link[x];
}
void play()
{
    V.clear();
    for(int i=1; i<=n; i++)
    {
        int cz=k-ai[i];
        if(ai[i]==0)continue;
        if(bi[ai[i]]==0)continue;
        bi[ai[i]]--;
        if(bi[ai[i]]==0)Link[ai[i]]=Link[ai[i]-1];
        if(cz<=0)continue;
        if(cz>10000)cz=10000;
        int f=Find(cz);
        bi[f]--;
        if(bi[f]==0)Link[f]=Link[f-1];
        V.push_back(f*ai[i]);
    }
    int p=V.size();
    LL ans=0;
    sort(V.begin(),V.end(),cmp);
    for(int i=0; i<min(p,m); i++)
    {
        ans+=V[i];
    }
    printf("%lld\n",ans);
}
int main()
{

    scanf("%d",&T);
    while(T--)
    {
        init();
        play();
    }
//    cout << "Hello world!" << endl;
    return 0;
}

import java.math.BigInteger;
import java.util.*;
public class zoj3903 {
	public static BigInteger pow2(long p) {
		BigInteger big = BigInteger.ZERO;
		BigInteger nn = new BigInteger(p + "");
		BigInteger nn1 = nn.add(BigInteger.ONE);
		BigInteger tw = new BigInteger(2 + "");
		BigInteger si = new BigInteger(6 + "");
		big = nn.multiply(nn1).multiply(nn.multiply(tw).add(BigInteger.ONE));
		return big.divide(si);
	}

	public static BigInteger pow3(long p) {
		BigInteger big = BigInteger.ZERO;
		BigInteger nn = new BigInteger(p + "");
		BigInteger nn1 = nn.add(BigInteger.ONE);
		BigInteger tw = new BigInteger(2 + "");
		big = nn.multiply(nn1).divide(tw);
		return big.multiply(big);
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int T;
		Scanner sca = new Scanner(System.in);
		T = sca.nextInt();
		long p;
		BigInteger MOD = new BigInteger("1000000007");
		long o = 1;
		BigInteger tw = new BigInteger(2 + "");
		for (int i = 1; i <= T; i++) {
			p = Long.parseLong(sca.next());
			p++;
			BigInteger q = new BigInteger((p - 1) + "");
			BigInteger b1 = pow3(p).subtract(pow3(o));
			BigInteger b2 = pow2(p).subtract(pow2(o));
			BigInteger ans = b1.subtract(b2);
			BigInteger b3 = pow2((p - 1) * 2).subtract(pow2(p));	
			b3 = b3.multiply(q);
			BigInteger b4 = pow3((p - 1) * 2).subtract(pow3(p));
			BigInteger b5 = pow2((p - 1) * 2).subtract(pow2(p));
			b5=b5.multiply(new BigInteger(p+""));
			BigInteger ans1 = b4.subtract(b5);
			ans1 = b3.subtract(ans1);
			
			ans = ans.add(ans1);
			
			ans =ans.add(pow2(p - 1).multiply(tw).multiply(tw));
			ans = ans.divide(tw);
			BigInteger st = q.multiply(new BigInteger(p + "")).divide(tw)
					.multiply(q).multiply(q);
			ans = ans.add(st);
			ans = ans.mod(MOD);
			System.out.println(ans);

		}

	}

}