Summer training #2

A:不管有没有负数一顿操作之后肯定只有正数又因为A=A-B 所以最大值是一直在减小的所以一定有结果

B:..一开始以为求min操作数 WA了2发直接求所有数的GCD如果所有数的GCD都不是1的话就无解有解就无脑扫2次就好了

C:给你一个char正方形矩阵要求a[i][j]四个正方形对称轴都一样可以任意change 直接爆搜八分之一块就行了.

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#define foror(i,a,b) for(i=a;i<b;i++)

#define foror2(i,a,b) for(i=a;i>b;i--)

#define EPS 1.0e-8

#define PI acos(-1.0)

#define INF 3000000000

#define MOD 1000000007

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define lson o<<1, l, m

#define rson o<<1|1, m+1, r

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

int cmp(int a,int b)

{

        return a<b;

}

char s[][];

int visit[][];

int gcd1(int a, int b)

{ return a ==  ? b : gcd1(b % a, a); }

map<char,int> mp;

int main()

{

 //freopen("in.txt", "r", stdin);

 //freopen("out.txt", "w", stdout);

 int t;

 cin >> t;

 while(t--)

 {

        int anser=;

        int n;

        char ch;

        cin >> n;

        int mid=(n+)/;

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

        mem(visit[i],);

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

        {

                scanf("%s",s[i]+);

        }

        /*for(int i=1;i<=n;i++)

        {

                printf("%s\n",s[i]+1);

        }*/

        int maxn=;

        for(int i=;i<=(n+)/;i++)

                for(int j=;j<=(n+)/;j++)

        {

              if(visit[i][j]==)

                continue;

              if(i==j&&i+j==n+)

               continue;

              if(i==j||i+j==n+||*i==n+||*j==n+)

              {

                int ans=;

                mp.clear();

                mp[s[i][j]]++;

                mp[s[i][n+-j]]++;

                mp[s[n+-i][j]]++;

                mp[s[n+-i][n+-j]]++;

                mp[s[j][i]]++;

                mp[s[n+-j][i]]++;

                mp[s[j][n+-i]]++;

                mp[s[n+-j][n+-i]]++;

                visit[i][j]=;

                visit[i][n+-j]=;

                visit[n+-i][j]=;

                visit[n+-i][n+-j]=;

                visit[j][i]=;

                visit[n+-j][i]=;

                visit[j][n+-i]=;

                visit[n+-j][n+-i]=;

                ans=max(ans,mp[s[i][j]]);

                ans=max(ans,mp[s[i][n+-j]]);

                ans=max(ans,mp[s[n+-i][j]]);

                ans=max(ans,mp[s[n+-i][n+-j]]);

                ans=max(ans,mp[s[j][i]]);

                ans=max(ans,mp[s[n+-j][i]]);

                ans=max(ans,mp[s[j][n+-i]]);

                ans=max(ans,mp[s[n+-j][n+-i]]);

                //cout <<i << ""<<j<<endl;

                //cout<<4-ans/2<<endl;

                anser+=-ans/;

              }

              else

              {

                int ans=;

                mp.clear();

                mp[s[i][j]]++;

                mp[s[i][n+-j]]++;

                mp[s[n+-i][j]]++;

                mp[s[n+-i][n+-j]]++;

                mp[s[j][i]]++;

                mp[s[n+-j][i]]++;

                mp[s[j][n+-i]]++;

                mp[s[n+-j][n+-i]]++;

                visit[i][j]=;

                visit[i][n+-j]=;

                visit[n+-i][j]=;

                visit[n+-i][n+-j]=;

                visit[j][i]=;

                visit[n+-j][i]=;

                visit[j][n+-i]=;

                visit[n+-j][n+-i]=;

                ans=max(ans,mp[s[i][j]]);

                ans=max(ans,mp[s[i][n+-j]]);

                ans=max(ans,mp[s[n+-i][j]]);

                ans=max(ans,mp[s[n+-i][n+-j]]);

                ans=max(ans,mp[s[j][i]]);

                ans=max(ans,mp[s[n+-j][i]]);

                ans=max(ans,mp[s[j][n+-i]]);

                ans=max(ans,mp[s[n+-j][n+-i]]);

                //cout <<i << ""<<j<<endl;

                //cout <<8-ans<<endl;

                anser+=-ans;

              }

        }

        cout << anser<<endl;

 }

  return ;

}

D:画人脸

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#define foror(i,a,b) for(i=a;i<b;i++)

#define foror2(i,a,b) for(i=a;i>b;i--)

#define EPS 1.0e-8

#define PI acos(-1.0)

#define INF 3000000000

#define MOD 1000000009

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define lson o<<1, l, m

#define rson o<<1|1, m+1, r

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

char ans[][][];

int len1[],len2[];

void init()

{

    for(int i=;i<;i++)

        for(int j=;j<;j++)

        ans[][i][j]='*';

     ans[][][]=ans[][][]=ans[][][]=' ';

     ans[][][]=ans[][][]=ans[][][]=' ';

     ans[][][]=ans[][][]=ans[][][]=' ';

     ans[][][]=ans[][][]=ans[][][]=' ';

     ans[][][]=ans[][][]=' ';

     ans[][][]=ans[][][]=' ';

     len1[]=,len2[]=;

     for(int i=;i<=;i++)

        len1[i]=len1[i-]*+;

     for(int i=;i<=;i++)

        len2[i]=len2[i-]*+;

}

void draw1(int x,int x1,int y1,int x2,int y2)//y2>y1 x2>x1

{

        if(x1==x2)

        {

        for(int i=y1;i<=y2;i++)

        {

        ans[x][x1][i]='*';

        }

        }

        else

        {

        for(int i=x1;i<=x2;i++)

        ans[x][i][y1]='*';

        }

}

void draw(int x,int len)

{

 int jian0=len>>;

 int jianer=len>>;

 int jianyi=len>>;

 draw1(x,,,,len-);

 draw1(x,,,len-,);

 draw1(x,,len-,len-,len-);

 draw1(x,len-,,len-,len-);

 //TS;

 draw1(x,jianer,jianer,jianer+jianyi-,jianer);

 draw1(x,jianer,jianer,jianer,jianer+jianyi-);

 draw1(x,jianer+jianyi,jianer,jianer+jianyi,jianer+jianyi-);

 draw1(x,jianer,jianer+jianyi-,jianer+jianyi,jianer+jianyi-);

 //TS;

 draw1(x,jianer,jianer+*jianyi-,jianer,jianer+*jianyi-);

 draw1(x,jianer,jianer+*jianyi-,jianer+jianyi,jianer+*jianyi-);

 draw1(x,jianer,jianer+*jianyi-,jianer+jianyi,jianer+*jianyi-);

 draw1(x,jianer+jianyi,jianer+*jianyi-,jianer+jianyi,jianer+*jianyi-);

 //TS;

 for(int i=len-;i>=jian0;i--)

 for(int j=jianyi+jian0-;j>=jianyi;j--)

 {

        ans[x][i][j]=ans[x-][len--i][jianyi+jian0--j];

 }

 //cout << x<<endl;

}

int main()

{

 //freopen("in.txt", "r", stdin);

 //freopen("out.txt", "w", stdout);

 int n;

 init();

 //TS;

 for(int x=;x<=;x++)

 {

        draw(x,<<x);

 }

 while(scanf("%d",&n)==)

 {

        if(n<)

        break;

        int num=log2(n);

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

        {

                for(int j=;j<n;j++)

                {

                        printf("%c",ans[num][i][j]);

                }

                cout <<endl;

        }

        cout <<endl;

 }

  return ;

}

E:先离散化离散化 +模拟双端队列

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;

struct Node{

    int c,p;

    bool operator<(const Node& rhs)const{

        if(c!=rhs.c) return c<rhs.c;

        return p<rhs.p;

    }

}s[];

int N,K;

int cal(int st,int ed){

    int k=,sum=,ret=;

    for(int p=st,q=st;q<=ed;){

        if(k<=K){

            sum++;

            ret=max(ret,sum);

            k+=s[q+].p-s[q].p-;

            q++;

        }else{

            sum--;

            k-=s[p+].p-s[p].p-;

            p++;

        }

    }

    return ret;

}

int main()

{

//    freopen("data.in","r",stdin);

    while(scanf("%d%d",&N,&K)==){

        int i,j;

        for(i=;i<N;s[i].p=i++)

            scanf("%d",&s[i].c);

        sort(s,s+N);

        int ans=;

        for(i=;i<N;i=j+){

            int c=s[i].c;

            j=i;

            while(j+<N && s[j+].c==c) j++;

            ans=max(ans,cal(i,j));

        }

        printf("%d\n",ans);

    }

    return ;

}

F:递推DP

1）用dp[i][j]表示源串第i次操作后，与目标串不同的位置有几个。

（2）对于dp[i][j]，有j个位置与目标串不同，所以可以将j个位置中的t个位置变为与目标串相同，把其余n-j个位置中的m-t个位置变为与目标串不同，转移方程就是dp[i+1][j-t+m-t]+=dp[i][j]*C(t,j)*C(m-t,n-j)；

（3）如果j+m>n，那么t就无法从0开始枚举了，所以此时t从（m-(n-j)）开始（m次操作中有n-j次操作落在n-j个相同位置的里面，还有m-(n-j)次操作落在j个不同位置的里面）。

（4）初始条件就是用cnt统计源串和目标串有多少个位置不一样，dp[0][cnt]=1，通过dp把路径都找出来，那么答案就是dp[k][0]。

#include <iostream>

#include <cstdio>

using namespace std;

typedef long long LL;

const LL MOD=;

LL dp[][],c[][];

char s[],d[];

void _init(){

    c[][]=;

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

        c[i][]=;

        for(int j=;j<=i;j++){

            c[i][j]=(c[i-][j-]+c[i-][j])%MOD;

        }

    }

}

int main()

{

    int n,k,m, i,j,t,x;

    _init();

    while(scanf("%d%d%d",&n,&k,&m)==){

        scanf("%s",s);

        scanf("%s",d);

        x=;

        for(i=;i<n;i++)

            if(s[i]!=d[i])x++;

        for(i=;i<=k;i++)

            for(j=;j<=n;j++)

            dp[i][j]=;

        dp[][x]=;

        for(i=;i<k;i++)//ith operation

            for(j=;j<=n;j++){//j different

                if(dp[i][j]==)continue;

                for(t=max(,m-(n-j));t<=j&&t<=m;t++)//j+m should no biger than n or at least (m-(n-j)) should be change

                    dp[i+][j-t+m-t]=(dp[i+][j-t+m-t]+dp[i][j]*c[j][t]%MOD*c[n-j][m-t]%MOD)%MOD;

            }

        LL ans=dp[k][];

        printf("%lld\n",ans);

    }

    return ;

}

G:给你一个完全/非完全二叉树求最小的宽度首先肯定把根节点的两个叶子放在同一侧是最优的然后让较宽的那一枝在最里面是最优的以一个节点为例:如果只有一个枝就无限地延伸 ans[x]=ans[a[x][0]] 如果有两个枝的话相同就宽度+1 不同就取最大的

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#define foror(i,a,b) for(i=a;i<b;i++)

#define foror2(i,a,b) for(i=a;i>b;i--)

#define EPS 1.0e-8

#define PI acos(-1.0)

#define INF 3000000000

#define MOD 1000000009

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define lson o<<1, l, m

#define rson o<<1|1, m+1, r

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

vector<vector<int> >a();

int ans[];

void doit(int x)

{

        if(a[x].size()==)

        {

        ans[x]=;

        }

        else if(a[x].size()==)

        {

        doit(a[x][]);

        ans[x]=ans[a[x][]];

        }

        else if(a[x].size()==)

        {

        doit(a[x][]);

        doit(a[x][]);

        if(ans[a[x][]]==ans[a[x][]])

        ans[x]=ans[a[x][]]+;

        else

        ans[x]=max(ans[a[x][]],ans[a[x][]]);

        }

}

int main()

{

 //freopen("in.txt", "r", stdin);

 //freopen("out.txt", "w", stdout);

 int n;

 while(scanf("%d",&n)==)

 {

        mem(ans,);

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

        a[i].clear();

        for(int i=;i<=n-;i++)

        {

        int x;

        scanf("%d",&x);

        a[x].push_back(i+);

        }

        doit();

        /*for(int i=1;i<=n;i++)

                printf("%d ",ans[i]);

        cout << endl;*/

        cout << ans[]<<endl;

 }

  return ;

}

H:几何

题目大意：给你一个三角形的内切圆半径跟外接圆半径，求解出符合条件的三角形，输出三角形的三条边的长度，如果没有符合条件的三角形，输出“NO Solution!”。

解题思路：这个题是SP,既是因为情况不唯一，而且还有精度的误差。

　　首先能够想到的就是NO Solution!的情况，即当内切圆半径等于1/2外接圆半径时，此时内切圆最大，而三角形为等边三角形，如图。

　　其次要解决的就是怎么构造三角形的问题，因为解不唯一，所以只要列举出一种解就OK，于是就很容易的想到构造等腰三角形，在最大与最小之间二分等腰三角形的底边长度，解三角形得到答案，如图。

#define sqr(x) (x)*(x)

int main()

{

    double r, R;

    while(~scanf("%lf%lf", &r, &R)){

        if(R < *r){

            printf("NO Solution!\n");

            continue;

        }

        double left = , right = sqrt(3.0)*R;

        while(right-left > eps){

            double mid = (left+right)/2.0;

            double t = sqrt(sqr(sqrt(sqr(R)-sqr(mid/2.0))+R)+sqr(mid/2.0));

            //printf("%lf\n", t);

            if(sqr(sqrt(sqr(r)+sqr(t-mid/2.0))+r) + sqr(mid/2.0) - sqr(t) < eps){

                right = mid;

            }

            else {

                left = mid;

            }

        }

        double p = sqrt(sqr(sqrt(sqr(R)-sqr(left/2.0))+R)+sqr(left/2.0));

        printf("%.16lf %.16lf %.16lf\n", p, p, left);

    }

    return ;

}

/*Sample test*/

/*

*/

ZOJ

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