本厂做了3个水体，被略哭

水题1  暴力乱搞

题目：

回文数猜想

27228
37649

3
27228--->109500--->115401--->219912
2
37649--->132322--->355553

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const  int maxn=1000000;

int n,path[maxn];
char str[maxn],xx[maxn],yy[maxn];

bool isprime(int n)
{
    itoa(n,str,10);
    int len=strlen(str);
    for(int i=0;i<len/2;i++)
    {
        if(str[i]!=str[len-i-1])
            return false;
    }
    return true;
}

int main()
{
    int ok,cnt,temp;
    while(~scanf("%d",&n))
    {
        cnt=0;
        path[++cnt]=n;
        ok=1;
        while(ok)
        {
            if(!isprime(n))
            {
                 int zz=0;
                 itoa(n,xx,10);
                 int len=strlen(xx);
                 for(int i=len-1;i>=0;i--)
                 {
                     yy[zz]=xx[i];
                     zz++;
                 }
                 yy[zz]=‘\0‘;
                 n=n+atoi(yy);
                 path[++cnt]=n;
            }
            else
                ok=0;
        }
        printf("%d\n",cnt-1);
        for(int i=1;i<cnt;i++)
            printf("%d--->",path[i]);
        printf("%d\n",path[cnt]);
    }
    return 0;
}

UVA10954 ADD ALL（优先队列）

能够说是哈弗曼模型吧

就是给n个数，然后求n个数的和的和最小

就是构造一个数越小优先级越高的队列。每次取前两位，然后求累加和就可以

题目：

Problem F
Add All
Input: standard input
Output: standard output

Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity to it.

Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add 1 and 10, you need a cost of11. If you want to add 1, 2 and 3. There are several ways –

I hope you have understood already your mission, to add a set of integers so that the cost is minimal.

Input

Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by N positive integers (all are less than 100000). Input is terminated by a case where the value of N is zero. This case should not be processed.

Output

For each case print the minimum total cost of addition in a single line.

Sample Input                           Output for Sample Input

Problem setter: Md. Kamruzzaman, EPS

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#include<vector>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=5000+10;

priority_queue<int,vector<int>,greater<int> >Q;

int a[maxn],n;

int main()
{
    int last,ans;
    while(~scanf("%d",&n),n)
    {
        while(!Q.empty()) Q.pop();
        ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&last);
            Q.push(last);
        }
        for(int i=1;i<n;i++)
        {
            int a=Q.top();Q.pop();
            int b=Q.top();Q.pop();
            last=a+b;
            ans+=last;
            Q.push(last);
        }
        printf("%d\n",ans);
    }
    return 0;
}

题目：

The plan of city rebuild

2
4 5
0 1 10
0 2 20
2 3 40
1 3 10
1 2 70
1 1
4 1 60
2
2 3
3 3
0 1 20
2 1 40
2 0 70
2 3
0 3 10
1 4 90
2 4 100
0

70
160

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
priority_queue<int,vector<int>,greater<int> >Q;


const int maxn=1000+10;
bool vis[maxn];
int cnt,yy,root[maxn];

struct Edge
{
    int u,v,w;
}edge[maxn*maxn];

bool cmp(Edge a,Edge b)
{
    return a.w<b.w;
}

int findroot(int x)
{
    if(root[x]!=x)
        root[x]=findroot(root[x]);
    return root[x];
}

int kruscal(int n)
{
    for(int i=0;i<maxn;i++)
        root[i]=i;
    int l,r,ans=0;
    int xx=0;
    for(int i=1;i<=cnt;i++)
    {
        int l=edge[i].u;
        int r=edge[i].v;
        int val=edge[i].w;
        if(vis[l]||vis[r])
            continue;
        else
        {
            int fx=findroot(l);
            int fy=findroot(r);
            if(fx==fy)  continue;
            else
            {
                root[fx]=fy;
               // printf("%d %d %d\n",l,r,val);
                ans+=val;
                xx++;
                if(xx==n-1)
                    return ans;
            }
        }
    }
    return -1;
}

int main()
{
    int t,n,m,sum,l,r,mid,newc,newr,old;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,false,sizeof(vis));
        cnt=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&l,&r,&mid);
            edge[++cnt].u=l;
            edge[cnt].v=r;
            edge[cnt].w=mid;
           // printf("%d %d %d\n",l,r,mid);
        }
        scanf("%d%d",&newc,&newr);
        for(int i=1;i<=newr;i++)
        {
            scanf("%d%d%d",&l,&r,&mid);
            edge[++cnt].u=l;
            edge[cnt].v=r;
            edge[cnt].w=mid;
          //  printf("%d %d %d\n",l,r,mid);
        }
        scanf("%d",&old);
        for(int i=1;i<=old;i++)
        {
            scanf("%d",&l);
            vis[l]=true;
        }
        sort(edge+1,edge+1+cnt,cmp);
        int ans=kruscal(n+newc-old);
        if(ans==-1)  printf("what a pity!\n");
        else printf("%d\n",ans);
    }
    return 0;
}

我居然看例子那个奇妙的字母是B，然后就以为不变了  

哎  直接枚举A--Z看哪个字母满足条件就可以，

题目：

破译password

30
17 6 9 8 3 0 1 6 7 4 5 10 11 8 9 14 15 12 13 18 19 16 17 22 23 20 21 26 27 24

SDKJABCDEFGHIJKLMNOPQRSTUVWXYZ

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
priority_queue<int,vector<int>,greater<int> >Q;


const int maxn=10000+10;

int a[maxn],n;

int main()
{
    char ch;
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<26;i++)
        {
            ch=i+‘A‘;
            for(j=1;j<=n;j++)
            {
                 if((a[j]^ch)<‘A‘||(a[j]^ch)>‘Z‘)
                     break;
            }
            if(j==n+1)
                break;
        }
        //printf("%c\n",ch);
        for(int i=1;i<n;i++)
            printf("%c",ch^a[i]);
        printf("%c\n",ch^a[i]);
    }
    return 0;
}

codeforces  475 B. Strongly Connected City

这个题目是求全图的联通，由于数据范围非常小。所以直接枚举全部的顶点，然后保搜得到全部的点。

最后看是否所有联通。我sb的把数组开小。导致调了一个中午，哎。郁闷啊  

题目链接：

huangjing

题目：

Imagine a city with n horizontal streets crossing m vertical streets, forming an (n?-?1)?×?(m?-?1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.

The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.

3 3
><>
v^v

NO

4 6
<><>
v^v^v^

YES

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
priority_queue<int,vector<int>,greater<int> >Q;

const int maxn=20+10;

struct Node
{
    int a,b;
}node[maxn*maxn];

bool vis[maxn*maxn][maxn*maxn];
char path[maxn];
int n,m;

void dfs(int xx,int yy)
{
    if(vis[xx][yy]) return;
    else
    {
        vis[xx][yy]=true;
        if(node[yy].a==0&&yy%m!=1)  dfs(xx,yy-1);
        else if(node[yy].a==1&&yy%m!=0)  dfs(xx,yy+1);
        if(node[yy].b==0&&yy>m)  dfs(xx,yy-m);
        else if(node[yy].b==1&&yy<(n*m-m+1))  dfs(xx,yy+m);
    }
}

int main()
{
    int ok;
    while(~scanf("%d%d",&n,&m))
    {
         ok=1;
         scanf("%s",path+1);
         for(int i=1;i<=n;i++)
             for(int j=1;j<=m;j++)
             {
                 if(path[i]==‘<‘)
                    node[m*(i-1)+j].a=0;
                 else
                    node[m*(i-1)+j].a=1;
             }
         scanf("%s",path+1);
         for(int j=1;j<=m;j++)
             for(int i=1;i<=n;i++)
             {
                 if(path[j]==‘^‘)
                    node[m*(i-1)+j].b=0;
                 else
                    node[m*(i-1)+j].b=1;
             }
        for(int i=1;i<=n*m;i++)
        {
            memset(vis,false,sizeof(vis));
            dfs(i,i);
            for(int j=1;j<=n*m;j++)
            {
                if(!vis[i][j])
                {
                    ok=0;
                    break;
                }
            }
            if(!ok)  break;
        }
        if(!ok)  puts("NO");
        else puts("YES");
    }
    return 0;
}