Pretty Matrix

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Time Limit: 1 Second      

Memory Limit: 65536 KB

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DreamGrid's birthday is coming. As his best friend, BaoBao is going to prepare a gift for him.

As we all know, BaoBao has a lot of matrices. This time he picks an integer matrix with  rows and  columns from his collection, but he thinks it's not pretty enough. On the one hand, he doesn't want to be stingy, but some integers in the matrix seem to be too small. On the other hand, he knows that DreamGrid is not good at algebra and hates large numbers, but some integers in the matrix seem to be too large and are not suitable for a gift to DreamGrid.

Based on the above consideration, BaoBao declares that a matrix is pretty, if the following conditions are satisfied:

1. For every integer  in the matrix, .
2. For every integer  in the matrix, .

where  is the integer located at the -th row and the -th column in the matrix, and  and  are two integers chosen by BaoBao.

Given the matrix BaoBao picks, along with the two integers  and , please help BaoBao change some integers in the matrix (BaoBao can change an integer in the matrix to any integer) so that the matrix becomes a pretty matrix. As changing integers in the matrix is tiring, please tell BaoBao the minimum number of integers in the matrix he has to change to make the matrix pretty.

Input

There are multiple test cases. The first line of input is an integer  (about 100), indicating the number of test cases. For each test case:

The first line contains four integers , ,  and . (). Their meanings are described above.

For the next  lines, the -th line contains  integers  (), representing the original matrix.

Output

For each test case output one line indicating the answer. If it's impossible to make the matrix pretty, print "No Solution" (without quotes) instead.

Sample Input

23 4 2 33 2 2 22 1 2 32 3 100 32 1 2 112

Sample Output

2No Solution

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#include 
     
      using namespace std;int main(){    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);    int T;    cin>>T;    while(T--)    {        int n,m,a,b,ans=0;        cin>>n>>m>>a>>b;        for(int i=0; i
      
       >x;                if(x
       
        b)ans++;            }        if(a>b)cout<<"No Solution\n";        else cout<
        
         <<"\n";    }    return 0;}
        
       
      
     

PPAP

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Time Limit: 1 Second      

Memory Limit: 65536 KB

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"I have a pen. I have an apple. Uh! Applepen."

"I have a pen. I have pineapple. Uh! Pineapplepen."

The above lyrics are taken from PPAP, a single by Pikotaro. It was released as a music video on YouTube on 25 August 2016, and has since become a viral video. As of October 2017, the official video has been viewed over 126 million times.

Let's view this song from a mathematical perspective. In the lyrics there actually hides a function , which takes two lowercased string  and  as the input and works as follows:

• First, calculate  ( here means string concatenation).
• Then, capitalize the first character of  to get .
• Make  as its output, and the function is done.

For example, we have PPAP("pen", "apple") = "Applepen", and PPAP("pen", "pineapple") = "Pineapplepen".

Given two lowercased strings  and , your task is to calculate .

Input

The first line of the input contains an integer  (about 100), indicating the number of test cases. For each test case:

The first and only line contains two strings  and  () separated by one space. It's guaranteed that both  and  consist of only lowercase English letters.

Output

For each test case output one line containing one string, indicating .

Sample Input

3pen applepen pineappleabc def

Sample Output

ApplepenPineapplepenDefabc

---

#include 
     
      using namespace std;int main(){    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);    int T;    cin>>T;    while(T--)    {        string s,c;        cin>>s>>c;        c[0]-=32;        cout<
      
       <
       <<"\n";    }    return 0;}
      
     

Mergeable Stack

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Time Limit: 2 Seconds      

Memory Limit: 65536 KB

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Given  initially empty stacks, there are three types of operations:

• 1 s v: Push the value  onto the top of the -th stack.

• 2 s: Pop the topmost value out of the -th stack, and print that value. If the -th stack is empty, pop nothing and print "EMPTY" (without quotes) instead.

• 3 s t: Move every element in the -th stack onto the top of the -th stack in order.

  Precisely speaking, denote the original size of the -th stack by , and the original size of the -th stack by . Denote the original elements in the -th stack from bottom to top by , and the original elements in the -th stack from bottom to top by .

  After this operation, the -th stack is emptied, and the elements in the -th stack from bottom to top becomes . Of course, if , this operation actually does nothing.

There are  operations in total. Please finish these operations in the input order and print the answer for every operation of the second type.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers  and  (), indicating the number of stacks and the number of operations.

The first integer of the following  lines will be  (), indicating the type of operation.

• If , two integers  and  (, ) follow, indicating an operation of the first type.
• If , one integer  () follows, indicating an operation of the second type.
• If , two integers  and  (, ) follow, indicating an operation of the third type.

It's guaranteed that neither the sum of  nor the sum of  over all test cases will exceed .

Output

For each operation of the second type output one line, indicating the answer.

Sample Input

22 151 1 101 1 111 2 121 2 133 1 21 2 142 12 12 12 12 13 2 12 22 22 23 73 1 23 1 33 2 12 12 22 32 3

Sample Output

13121110EMPTY14EMPTYEMPTYEMPTYEMPTYEMPTYEMPTY

 

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卡函数，要用list的splice

就是这个合并很耗时间，用链表去模拟也是可以的

#include 
     
      using namespace std;const int N=3e5+5;list
      
       L[N];int main(){    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);    int T;    cin>>T;    while(T--)    {        for(int i=0; i
       
        >n>>q;        while(q--)        {            int op;            cin>>op;            if(op==1)            {                int a,b;                cin>>a>>b;                L[a].push_back(b);            }            else if(op==2)            {                int a;                cin>>a;                if(L[a].size())                {                    cout<
        
         <<"\n";                    L[a].pop_back();                }                else cout<<"EMPTY\n";            }            else            {                int a,b;                cin>>a>>b;                L[a].splice(L[a].end(),L[b]);            }        }    }    return 0;}
        
       
      
     

Traffic Light

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Time Limit: 1 Second      

Memory Limit: 131072 KB

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DreamGrid City is a city with  intersections arranged into a grid of  rows and  columns. The intersection on the -th row and the -th column can be described as , and two intersections  and  are connected by a road if .

At each intersection stands a traffic light. A traffic light can only be in one of the two states: 0 and 1. If the traffic light at the intersection  is in state 0, one can only move from  to  or ; If the traffic light is in state 1, one can only move from  to  or  (of course, the destination must be another intersection in the city).

BaoBao lives at the intersection , and he wants to visit his best friend DreamGrid living at the intersection . After his departure, in each minute the following things will happen in order:

• BaoBao moves from his current intersection to another neighboring intersection along a road. As a law-abiding citizen, BaoBao has to obey the traffic light rules when moving.
• Every traffic light changes its state. If a traffic light is in state 0, it will switch to state 1; If a traffic light is in state 1, it will switch to state 0.

As an energetic young man, BaoBao doesn't want to wait for the traffic lights, and he must move in each minute until he arrives at DreamGrid's house. Please tell BaoBao the shortest possible time he can move from  to  to meet his friend, or tell him that this is impossible.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers  and  (), indicating the size of the city.

For the following  lines, the -th line contains  integers  (), where  indicates the initial state of the traffic light at intersection .

The next line contains four integers , ,  and  (, ), indicating the starting intersection and the destination intersection.

It's guaranteed that the sum of  over all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the shortest possible time (in minute) BaoBao can move from  to  without stopping. If it is impossible for BaoBao to arrive at DreamGrid's house, print "-1" (without quotes) instead.

Sample Input

42 31 1 00 1 01 3 2 12 31 0 01 1 01 3 1 22 21 01 01 1 2 21 20 11 1 1 1

Sample Output

35-10

Hint

For the first sample test case, BaoBao can follow this path: .

For the second sample test case, due to the traffic light rules, BaoBao can't go from  to  directly. Instead, he should follow this path: .

For the third sample test case, it's easy to discover that BaoBao can only go back and forth between  and .

 按照要求dfs吧

#include 
     
      using namespace std;struct T{    int x,y,f;} tt;int dir[2]= {-1,1};int solve(){    int n,m;    scanf("%d%d",&n,&m);    int a[n+5][m+5],vis[n+5][m+5];    for(int i=1; i<=n; i++)        for(int j=1; j<=m; j++)            scanf("%d",&a[i][j]),vis[i][j]=0;    int x,y,s,t,f=0;    scanf("%d%d%d%d",&x,&y,&s,&t);    queue
      
        q;    vis[x][y]=1,q.push({x,y,f});    while(!q.empty())    {        tt=q.front();        q.pop();        if(tt.x==s&&tt.y==t) return tt.f;        int op=(a[tt.x][tt.y]+tt.f)%2;        if(op==1)        {            for(int i=0; i<2; i++)            {                x=tt.x,y=tt.y+dir[i],f=tt.f+1;                if(vis[x][y]||y<1||y>m) continue;                vis[x][y]=1,q.push({x,y,f});            }        }        else        {            for(int i=0; i<2; i++)            {                x=tt.x+dir[i],y=tt.y,f=tt.f+1;                if(vis[x][y]||x<1||x>n) continue;                vis[x][y]=1,q.push({x,y,f});            }        }    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)printf("%d\n",solve());    return 0;}