# ECUST 09年 校赛个人赛第六，七场总结

校赛个人赛第六，七场总结

这两场比赛体现了英文水平的重要性

第六场的题目超长，用词还诡异，话了很长时间才看懂

这两场题目都比较有难度，第六场我只出了2题

## A Grey Area

A题是很诡异的统计，是一道纯模拟就能过的题，其他的不多说了

代码：

```cpp
#include<iostream>
using namespace std;
int pe[20];
int main()
{
    int n,w;
    while(cin>>n>>w, n || w)
    {
        int value;
        int peNum = 0;
        int Max = 0;
        int valueMax = 0;
        memset(pe,0,sizeof(pe));
        while(n --)
        {
            scanf("%d",&value);
            while(value >= Max)
            {
                peNum ++;
                Max = peNum * w;
            }
            pe[value / w] ++;
            if(pe[value / w] > valueMax)
                valueMax = pe[value / w];
        }

        double output = 0.01;
        for(int i = 0 ; i < peNum ; i ++)
            output += ((double)pe[i] / (double)valueMax)
                * ((double)(peNum - i - 1) / (double)(peNum - 1));

        printf("%lf\n",output);
    }
    return 0;
}
```

## B Expected Allowance

B题和A题类似，也是纯暴力

代码：

```cpp
#include<iostream>
#include<cmath>
using namespace std;
long value[10000];

void cal(long now,int pos,int n,int m)
{
    for(int i = 1 ; i <= m ; i ++)
        if(pos == n)
            value[now + i] ++;
        else
            cal(now + i,pos + 1,n,m);
}


int main()
{
    long m,n,k;
    while(scanf("%ld %ld %ld",&n,&m,&k), n || m || k )
    {
        memset(value,0,sizeof(value));
        cal(0,1,n,m);
        double output = 0.0;
        double fm = pow((double)m,(double)n);
        for(long i = n ; i <= n * m ; i ++)
            if(i - k <= 0)
                output += (double)value[i] / fm;
            else
                output += (double)value[i] / fm * (double)(i - k);

        printf("%.8lf\n",output);
    }
    return 0;
}
```

## 第七场

第七场很多题目都有想法，可是很多都没能写出来

同样B和E是水题，直接暴力能过，不过就是要有耐性

## B要注意一下判断的优先级.

代码如下：

```cpp
#include<iostream>
#include<cstring>
using namespace std;
double bank[10][10000];
char cmd[20];
int isExist[10][10000];
int main()
{
    int n = 1;
    //freopen("b.in","r",stdin);
    while(scanf("%d",&n),n)
    {
        memset(bank,0,sizeof(bank));
        memset(isExist,0,sizeof(isExist));
        while(n --)
        {
            int tmpInt1,tmpInt2;
            double tmpMoney;
            scanf("%d/%d %lf",&tmpInt1,&tmpInt2,&tmpMoney);
            isExist[tmpInt2][tmpInt1] = 1;
            bank[tmpInt2][tmpInt1] = tmpMoney;
        }
        while(scanf("%s",cmd) , strcmp(cmd,"end"))
        {
            if(!strcmp(cmd,"withdraw"))
            {
                int tmpInt1,tmpInt2;
                double tmpMoney;
                scanf("%d/%d %lf",&tmpInt1,&tmpInt2,&tmpMoney);
                printf("withdraw %.2lf: ",tmpMoney);
                if(isExist[tmpInt2][tmpInt1])
                {
                    if(bank[tmpInt2][tmpInt1] < tmpMoney)
                        printf("insufficient funds\n");
                    else
                    {
                        printf("ok\n");
                        bank[tmpInt2][tmpInt1] -= tmpMoney;
                    }
                }
                else
                    printf("no such account\n");
            }
            else if(!strcmp(cmd,"deposit"))
            {
                int tmpInt1,tmpInt2;
                double tmpMoney;
                scanf("%d/%d %lf",&tmpInt1,&tmpInt2,&tmpMoney);
                printf("deposit %.2lf: ",tmpMoney);
                if(isExist[tmpInt2][tmpInt1])
                {
                    printf("ok\n");
                    bank[tmpInt2][tmpInt1] += tmpMoney;
                }
                else
                    printf("no such account\n");
            }
            else if(!strcmp(cmd,"create"))
            {
                int tmpInt1,tmpInt2;
                scanf("%d/%d",&tmpInt1,&tmpInt2);
                printf("create: ");
                if(isExist[tmpInt2][tmpInt1])
                    printf("already exists\n");
                else
                {
                    printf("ok\n");
                    isExist[tmpInt2][tmpInt1] = 1;
                    bank[tmpInt2][tmpInt1] = 0;
                }
            }
            else if(!strcmp(cmd,"transfer"))
            {
                int tmpInt1,tmpInt2,tmpInt3,tmpInt4;
                double tmpMoney;
                scanf("%d/%d %d/%d %lf",&tmpInt1,&tmpInt2,&tmpInt3,&tmpInt4,&tmpMoney);
                printf("transfer %.2lf: ",tmpMoney);
                if(isExist[tmpInt2][tmpInt1] && isExist[tmpInt4][tmpInt3])
                {
                    if(tmpInt2 == tmpInt4 && tmpInt1 == tmpInt3)
                        printf("same account\n");
                    else if(bank[tmpInt2][tmpInt1] < tmpMoney)
                        printf("insufficient funds\n");
                    else
                    {
                        if(tmpInt2 == tmpInt4)
                            printf("ok\n");
                        else
                            printf("interbank\n");
                        bank[tmpInt2][tmpInt1] -= tmpMoney;
                        bank[tmpInt4][tmpInt3] += tmpMoney;
                    }
                }
                else
                    printf("no such account\n");
            }
        }
        printf("end\n\n");
    }
    printf("goodbye");
    return 0;
}
```

## E Stock Exchange

我这题TLE一次，原因是很郁闷的忘了把freopen函数给注释掉 -\_-!!

我的代码如下:

```cpp
#include<iostream>
#include<cstring>
using namespace std;
char Issuer[12];
struct Member
{
    char name[30];
    double money;
    char type[10];
};
Member mem[1001];
int main()
{
    int n;
    //freopen("e.txt","r",stdin);
    while(scanf("%d %s",&n,Issuer), strcmp(Issuer,"END") || n)
    {
        int i,j;
        for(i = 0 ; i < n ; i ++)
            scanf("%s %s %lf",mem[i].name,mem[i].type,&mem[i].money);

        printf("%s\n",Issuer);
        int num = 0;
        for(i = 0 ; i < n ; i ++)
        {
            num = 0;
            printf("%s:",mem[i].name);
            for(j = 0 ; j < n ; j ++)
            {
                if(mem[i].type[0] == mem[j].type[0])
                    continue;
                if(mem[i].type[0] == 'b'
                    && mem[j].money <= mem[i].money)
                    printf(" %s",mem[j].name),num ++;
                else if(mem[i].type[0] == 's'
                    && mem[j].money >= mem[i].money)
                    printf(" %s",mem[j].name),num ++;
            }
            if(num)
                putchar('\n');
            else
                printf(" NO-ONE\n");
        }
    }
    return 0;
}
```

## C题很烦，但是我觉得可以通过列出各种特殊情况出来死套。太烦了所以没写

## D题我觉得可以通过DP+Hash来完成，按最长子序列的反向DP（不过我觉得等我想出来DP方案时间也到了，就没写）

## F题我没什么思路

## G题我认为可以DP，主要是算出三个子段的最小值，且三个子段的长度和为k-1，但是DP的话要写24种分支，我觉得这个方法并不好，所以也没写

以上各题希望大牛能分享思路，指点一二。


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://blog.owent.net/2009/85.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.