SDUT 2012春季ACM内部测试赛5

http://acm.sdut.edu.cn/web/contest_show.php?contest_id=1097

A题：（字符串问题）就是给定字典。然后查找错误。分三种情况

1：One letter is missing (e.g., letter is written leter)or too much (e.g., letter is writtenlettter).

2：One letter is wrong (e.g., letter is written ketter)

3：The order of two adjacent letters is wrong (e.g., letter is written lettre)

然后判断。。之前做过类似的一道题目，1Y..比赛时，队长负责的题目，也是果断1Y..厉害。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <time.h>
#define maxn 10007
using namespace std;

char str[maxn][107];

bool isok(char *s1,char *s2)
{
    //printf("%s %s\n",s1,s2);
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int i,j;
    char h1[107] = {'0'};
    char h2[107] = {'0'};
    //相等时
    if (len1 == len2)
    {
        int l = 0;
        for (i = 0; i < len1; ++i)
        {
            if (s1[i] != s2[i])
            {
                h1[l] = s1[i];
                h2[l] = s2[i];
                l++;
            }
        }
        if (l == 1)//写错一个字母
        return true;
        else if (l == 2)//两个字母互换了位置
        {
            if (h1[0] == h2[1] && h1[1] == h2[0])
            return true;
            else
            return false;
        }
        else return false;
    }
    //len2<len1少写字母了
    else if (len1 == len2 + 1)
    {
        i = 0; j = 0;
        while (i < len1 && s1[i] == s2[j] && j < len2)
        {
           i++; j++;
        }
        i++;
        while (i < len1 && j <len2)
        {
            if (s1[i] != s2[j])
            return false;
            i++; j++;
        }
        return true;
    }
    //多写字母了
    else if (len1 + 1 == len2)
    {
        i = 0; j = 0;
        while (j < len2 && s1[i] == s2[j] && i < len1)
        {
           i++; j++;
        }
        j++;
        while (i < len1 && j <len2)
        {
            if (s1[i] != s2[j])
            return false;
            i++; j++;
        }
        return true;
    }
    return false;
}
int main()
{
    //freopen("d.txt","r",stdin);
    int n,i,q;
    char ch[107];
    scanf("%d",&n);
    for (i =0; i < n; ++i)
    scanf("%s",str[i]);
    //for (i =0; i < n; ++i)
   //printf(">>%s\n",str[i]);
    scanf("%d",&q);
    while (q--)
    {
        scanf("%s",ch);
        //printf(">>%s\n",ch);
        bool flag = false;
        //在字典中
        for (i = 0; i < n; ++i)
        {
            if (!strcmp(str[i],ch))
            {
                flag = true;
                break;
            }
        }
        if (flag)
        {
             printf("%s is correct\n",ch);
             continue;
        }
        //是否是拼写错误
        for (i = 0; i < n; ++i)
        {
            if (isok(str[i],ch))
            {
                flag = true;
                printf("%s is a misspelling of %s\n",ch,str[i]);
                break;
            }
        }
        //否则
        if (!flag)
        {
            printf("%s is unknown\n",ch);
        }
    }
    return 0;
}

F：果断的奇葩数据啊。。。。。。自己用hash挂链处理的哈希冲突。。。才开始使用的动态new开的内存。果断tle最后换成静态分配内存之后就A了。。。。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <time.h>
#define maxn 1000007
using namespace std;

struct node
{
    int num;
    int val;
    node *next;
}*p[maxn];
node H[maxn];
int pos;
int n,ct,a,b;

bool isok(int x,int y)
{
    node *t;
    bool flag = false;
    for (t = p[x]; t != NULL; t = t->next)
    {
        if (t->num == y)
        {
           t->val++;
           if (t->val == 2) ct++;
           //printf(">>%d %d\n",t->num,t->val);
           if (t->val == 3) return false;//遇到两次的直接跳出
           flag = true;
           break;
        }
    }
    //puts("DDDD");
    if (!flag)
    {
        node *q;
        q = &H[pos++];//静态分配内存，果断能过。。动态tle
        q->num = y; q->val = 1;
        q->next = p[x];
        p[x] = q;
    }
    return true;
}
int main()
{
    //freopen("d.txt","r",stdin);
    //int s = clock();
    while (~scanf("%d",&n))
    {
        if (!n) break;
        pos = 0;
        memset(p,0,sizeof(p));
        ct = 0;//记录出现两次的个数
        scanf("%d%d",&a,&b);
        int yu = b%n;
        while (1)
        {
            int tmp = yu%maxn;
            //printf(">>%d\n",tmp);
            if (!isok(tmp,yu))
            break;
            __int64 l = yu%n;
            yu = (((a%n)*((l*l)%n))%n + (b%n))%n;
        }
        printf("%d\n",n - ct);
    }
    //int e = clock();
    //printf(">>%d\n",e-s);
    return 0;
}

G: 这是我负责的一道题目，一个简单的贪心，我先是按+ - 数排序一下，同号的按序号从小到大排序，因为这样可以尽量让距离比较近的先进行交易。。开始没注意到long long 的提示果断的WA了一次。。后来von提醒了我一下才发现，改过了就A了。。

View Code

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define maxn 1000007
using namespace std;

struct node
{
    int val;
    int num;
}p[maxn];
int n;

int cmp(node a,node b)
{
    if ((a.val > 0 && b.val >0) || (a.val < 0 && b.val < 0))
    {
        return a.num < b.num;
    }
    else
    return a.val < b.val;
}
int main()
{
    int i,j,pos;
    while (cin>>n)
    {
        if (!n) break;
        for (i = 0; i < n; ++i)
        {
            cin>>p[i].val;
            p[i].num = i;
        }
        sort(p,p + n,cmp);//排序
        for (i = 0; i < n; ++i)
        {
            if (p[i].val >= 0)
            {
                pos = i;break;
            }
        }
        i = 0; j = pos;//从大于0的开始进行交易
       long long ct = 0;
        while (i < pos && j < n)
        {
            int tmp = p[j].val + p[i].val;
            //printf(">>%d %d\n",tmp,ct);
            if (tmp > 0)
            {
                long long m = abs(p[j].num - p[i].num) - 1;
                ct = ct + (-p[i].val) + (m)*(-p[i].val);
                p[j].val = tmp;
                p[i].val = 0;
                 i++;
            }
            else if (tmp == 0)
            {
                long long m = abs(p[j].num - p[i].num) - 1;
                ct = ct + (-p[i].val) + (m)*(-p[i].val);
                p[j].val = p[i].val = 0;
                i++; j++;
            }
            else if (tmp < 0)
            {
                long long m = abs(p[j].num - p[i].num) - 1;
                ct = ct + (p[j].val) + (m)*(p[j].val);
                p[j].val = 0;
                p[i].val = tmp;
                j++;
            }
        }
        printf("%I64d\n",ct);
    }
    return 0;
} 

F:本来该我负责的，可是我在G题上磨叽了很长时间，von就给做了。。。这道题好像是随机算法的问题，看了下随机算法做的可是做了之后是WA无语了。最后是分别判断没哥四边形，只要存在两个对角线和不相等。。肯定not homogeneous。。。

View Code

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define maxn 1007
using namespace std;

int a[maxn][maxn];
int main()
{
    int n,i,j;

    while (~scanf("%d",&n))
    {
        if (!n) break;
        bool flag = false;
        for (i = 0; i < n; ++i)
        {
            for (j = 0; j < n; ++j)
            {
                scanf("%d",&a[i][j]);
            }
        }
        for (i = 0; i < n - 1; ++i)
        {
            for (j = 0; j < n - 1; ++j)
            {
                if (a[i][j] + a[i + 1][j + 1] != a[i][j + 1] + a[i + 1][j])
                {
                    flag = true;
                    break;
                }
            }
            if (flag) break;
        }
        if (flag) printf("not homogeneous\n");
        else      printf("homogeneous\n");
    }
    return 0;
}