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:

#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:

#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:

#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:

#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。