2019 寧夏網絡賽

時間 2019-11-22

原文原文鏈接

A、Maximum Element In A Stack

As an ACM-ICPC newbie, Aishah is learning data structures in computer science. She has already known that a stack, as a data structure, can serve as a collection of elements with two operations:html

push, which inserts an element to the collection, and
pop, which deletes the most recently inserted element that has not yet deleted.

Now, Aishah hopes a more intelligent stack which can display the maximum element in the stack dynamically. Please write a program to help her accomplish this goal and go through a test with several operations.node

Aishah assumes that the stack is empty at first. Your program will output the maximum element in the stack after each operation. If at some point the stack is empty, the output should be zero.ios

Input

The input contains several test cases, and the first line is a positive integer web

To avoid unconcerned time consuming in reading data, each test case is described by seven integers $\le n \le 5 \times 10^6),p,q,m(1 \le p,q,m \le 10^9),SA,SB \ and \ SC(10^4 \le SA,SB,SC \le 10^6)n(1≤n≤5×106),p,q,m(1≤p,q,m≤109),SA,SB and SC(104≤SA,SB,SC≤106). Theinteger nnis the number of operations, and your program should generate all operations using the following code in C++.$ api

       
        int n, p, q, m; 
       
        unsigned int SA, SB, SC; 
       
unsigned int rng61() {
    SA ^= SA << 16;
    SA ^= SA >> 5;
    SA ^= SA << 1;
    unsigned int t = SA; SA = SB;
    SB = SC;
    SC ^= t ^ SA;
    return SC;
}
void gen(){
    scanf("%d%d%d%d%u%u%u", &n, &p, &q, &m, &SA, &SB, &SC);
    for(int i = 1; i <= n; i++) {
        if(rng61() % (p + q) < p)
            PUSH(rng61() % m + 1);
        else
            POP();
    }
}

The procedure PUSH(v) used in the code inserts a new element with value v into the stack and the procedure POP( ) pops the topmost element in the stack or does nothing if the stack is empty.app

Output

For each test case, output a line containing Case #x: y, where $\oplus^n_{i=1} (i \cdot a_i)⊕i=1n(i⋅ai) where a_iai is the answer after the ii-th operation and \oplus⊕ means bitwise xor.$ less

輸出時每行末尾的多餘空格，不影響答案正確性eclipse

樣例輸入

2
4 1 1 4 23333 66666 233333
4 2 1 4 23333 66666 233333

樣例輸出

Case #1: 19
Case #2: 1

樣例解釋

The first test case in the sample input has ide

POP();
POP();
PUSH(1);
PUSH(4).

The second test case also has 測試

PUSH(2);
POP();
PUSH(1);
POP().

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#include<stack>
#define mx 0x3f3f3f3f
#define ll long long
#define MAXN 100
using namespace std;
stack<ll>s;
int n,p,q,m;
ll ans;
unsigned int A, B, C;
unsigned int rng61()
{
    A ^= A << 16;
    A ^= A >> 5;
    A ^= A << 1;
    unsigned int t = A;
    A = B;
    B = C;
    C ^= t ^ A;
    return C;
}
void gen()
{
    scanf("%d%d%d%d%u%u%u",&n,&p,&q,&m,&A,&B,&C);
    for(int i=1;i<=n;i++)
    {
        if(rng61()%(p+q)<p)
        {
            ll temp=rng61()%m+1;
            if(s.size()==0)
                s.push(temp);
            else
                s.push(max(temp,s.top()));
        }
        else
        {
            if(s.size())
                s.pop();
        }
        if(s.size())
            ans^=i*s.top();
    }
} 


int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        ans=0;
        while(s.size())
            s.pop();
        gen();
        cout<<"Case #"<<i<<": "<<ans<<endl;
    }
    return 0;
}

C、Caesar Cipher

In cryptography, a Caesar cipher, also known as the shift cipher, is one of the most straightforward and most widely known encryption techniques. It is a type of substitution cipher in which each letter in the plaintext is replaced by a letter some fixed number of positions up (or down) the alphabet.

For example, with the right shift of

Now you have a plaintext and its ciphertext encrypted by a Caesar Cipher. You also have another ciphertext encrypted by the same method and are asked to decrypt it.

Input

The input contains several test cases, and the first line is a positive integer

For each test case, the first line contains two integers $\le n, m \le 50)m(1≤n,m≤50) indicating the length of the first two texts (a plaintext and its ciphertext) and the length of the third text which will be given. Each of the second line and the third line contains a string only with capital letters of length nn, indicating a given plaintext and its ciphertext respectively. The fourth line gives another ciphertext only with capital letters of length mm.$

We guarantee that the pair of given plaintext (in the second line) and ciphertext (in the third line) is unambiguous with a certain Caesar Cipher.

Output

For each test case, output a line containing Case #x: T, where

輸出時每行末尾的多餘空格，不影響答案正確性

樣例輸入

1
7 7
ACMICPC
CEOKERE
PKPIZKC

樣例輸出

Case #1: NINGXIA

題意：第一行和第二行給你一個翻譯樣例（明文和密文），讓你找翻譯規律，第三行給密文，要你按照翻譯規律找明文

//密碼解密
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
#define MAXN 100
using namespace std;
string s,ss;
int main()
{
    int n,m,t,k=0;
    cin>>t;
    while(t--)
    {
        k++;
        cin>>n>>m;
        cin>>s>>ss;
        int x=s[0]-ss[0];
        cin>>ss;
        cout<<"Case #"<<k<<": ";
        for(int i=0;ss[i];i++)
            printf("%c",( ss[i] -'A' + x+26  )% 26 +65);
        printf("\n");
    }
    return 0;
}
/*
25 5
UVWXYZ
NOPQRX
TUVWXY
*/

F、Moving On

Firdaws and Fatinah are living in a country with

Firdaws's home locates in the city

Input

The input contains several test cases, and the first line is a positive integer

For each test case, the first line contains two integers $\le n \le 200)n(1≤n≤200) which is the number of cities, and q (1 \le q \le 2 \times 10^4)q(1≤q≤2×104) which is the number of queries that will be given. The second line contains nn integers r_1, r_2, \cdots , r_nr1,r2,⋯,rn indicating the risk of kidnapping or robbery in the city 11 to nn respectively. Each of the following nn lines contains nn integers, the jj-th one in the ii-th line of which, denoted by d_{i,j}di,j, is the distance from the city iito the city jj.$

Each of the following

We guarantee that $\le r_i \le 10^5, 1 \le d_{i,j} \le 10^5 (i \neq j), d_{i,i} = 01≤ri≤105,1≤di,j≤105(i=j),di,i=0 and d_{i,j} = d_{j,i}di,j=dj,i. Besides, each query satisfies 1 \le u,v \le n1≤u,v≤n and 1 \le w \le 10^51≤w≤105.$

Output

For each test case, output a line containing Case #x: at first, where

輸出時每行末尾的多餘空格，不影響答案正確性

樣例輸入

樣例輸出

Case #1:
0
1
3
0
1
2

題意：輸入t，表示t個測試樣例
輸入n,q,表示n個點，q次詢問
下一行n個數表示[1,n]個點的危險值
接下來n行，每行3個數描述一條邊，點x到y的距離爲z;
最後q行，每行3個數，求從起點U到終點V,在每個點危險值不超過W的前提下的最短距離

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
#define MAXN 100
using namespace std;
int dp[250][250][250];//dp[k][i][j]表示 添加(第1~k小的危險值的城市後)的 i->j 最短路
int vis[250],rk[250];
bool cmp(int i,int j)
{
    return rk[i]<rk[j];
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int tt=1;tt<=t;tt++)
    {
        memset(dp,mx,sizeof(dp));
        int n,q;
        scanf("%d%d",&n,&q);
        for(int i=1;i<=n;i++)
        {
            vis[i]=i;//初始化排名
            scanf("%d",&rk[i]);
        }
        sort(vis+1,vis+n+1,cmp);//把城市的編號按風險等級從小到大排序，vis[排名]=編號
        // for(int i=1;i<=n;i++)
        //     cout<<vis[i]<<' ';
        // cout<<"<-----"<<endl;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&dp[0][i][j]);//初始化每一條邊的危險等級爲0
        for(int k=1;k<=n;k++)
        {
            int now=vis[k];
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                    dp[k][i][j]=min(dp[k-1][i][j],dp[k-1][i][now]+dp[k-1][now][j]); 
            }

        }
        printf("Case #%d:\n",tt);
        for(int i=1;i<=q;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            int k=0;
            for(int j=1;j<=n;j++)
                if(rk[vis[j]]<=w)//把危險值最接近w的邊都加進去以後的最短路
                    k=j;
            printf("%d\n",dp[k][u][v]);
        }
    }
    return 0;
}

H、Fight Against Monsters

It is my great honour to introduce myself to you here. My name is Aloysius Benjy Cobweb Dartagnan Egbert Felix Gaspar Humbert Ignatius Jayden Kasper Leroy Maximilian. As a storyteller, today I decide to tell you and others a story about the hero Huriyyah, and the monsters.

Once upon a time, citizens in the city were suffering from $HP_iHPi, and its attack value was ATK_iATKi.$

They fought in a cave through a turn-based battle. During each second, the hero Huriyyah was attacked by monsters at first, and the damage was the sum of attack values of all alive monsters. Then he selected a monster and attacked it. The monster would suffer the damage of

Now, my smart audience, can you calculate the minimum amount of total damages our hero should suffer before he won the battle?

Input

The input contains several test cases, and the first line is a positive integer $10^3103.$

For each test case, the first line contains an integers $\le n \le 10^5)n(1≤n≤105) which is the number of monsters. The ii-th line of the following nn lines contains two integers HP_iHPi and ATK_i (1 \le HP_i, ATK_i \le 10^5)ATKi(1≤HPi,ATKi≤105) which describe a monster.$

We guarantee that the sum of $10^6106.$

Output

For each test case, output a line containing Case #x: y, where

輸出時每行末尾的多餘空格，不影響答案正確性

樣例輸入

樣例輸出

Case #1: 19
Case #2: 14


題意：獵人打怪獸，怪獸有一個生命值和攻擊值，獵人的攻擊值在打不一樣的怪獸時從1開始隨攻擊次數遞增
攻擊順序時全部怪獸先攻擊獵人一次，獵人在選擇一個怪獸進行攻擊
問：獵人殺死全部怪獸，最少須要多少生命值

題解：貪心排序，按照每一個怪獸對獵人形成的總傷害（總傷害=攻擊力*攻擊次數）從大到小進行排序，獵人先把總傷害高的怪獸殺死

//貪心打怪
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#include<stack>
#define mx 0x3f3f3f3f
#define ll long long
#define MAXN 100
using namespace std;
struct node
{
    ll hp;
    ll ak;
    ll cnt;
}p[100005];
bool cmp(node a,node b)
{
    return a.ak*b.cnt>b.ak*a.cnt;
}
ll check(ll x)
{
    ll temp=x,num=0;
    for(int i=1;i<=temp;i++)
    {
        if(x>0)
            num++;
        else
            break;
        x=x-i;
    }
    return num;
}
int main()
{
    ll t,n,sum;
    scanf("%lld",&t);
    for(int j=1;j<=t;j++)
    {
        sum=0;
        scanf("%lld",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lld%lld",&p[i].hp,&p[i].ak);
            p[i].cnt=check(p[i].hp);
            sum=sum+p[i].ak;
        }
        sort(p,p+n,cmp);
        ll ans=0;
        for(int i=0;i<n;i++)
        {
            ans=ans+p[i].cnt*sum;
            sum=sum-p[i].ak;
        }
        printf("Case #%d: %lld\n",j,ans);
    }
}