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wangs — Sun, 22 Jul 2012 02:02:00 GMT

Problem Statement

We have a string originalWord. Each character of originalWord is either 'a' or 'b'. Timmy claims that he can convert it to finalWord using exactly k moves. In each move, he can either change a single 'a' to a 'b', or change a single 'b' to an 'a'.

You are given the strings originalWord and finalWord, and the int k. Determine whether Timmy may be telling the truth. If there is a possible sequence of exactly k moves that will turn originalWord into finalWord, return "POSSIBLE" (quotes for clarity). Otherwise, return "IMPOSSIBLE".

Definition

Class:	EasyConversionMachine
Method:	isItPossible
Parameters:	string, string, int
Returns:	string
Method signature:	string isItPossible(string originalWord, string finalWord, int k)
(be sure your method is public)

Notes

Timmy may change the same letter multiple times. Each time counts as a different move.

Constraints

originalWord will contain between 1 and 50 characters, inclusive.

finalWord and originalWord will contain the same number of characters.

Each character in originalWord and finalWord will be 'a' or 'b'.

k will be between 1 and 100, inclusive.

Examples

"aababba"

"bbbbbbb"

Returns: "IMPOSSIBLE"

It is not possible to reach finalWord in fewer than 4 moves.

"aabb"

"aabb"

Returns: "IMPOSSIBLE"

The number of moves must be exactly k=1.

"aaaaabaa"

"bbbbbabb"

Returns: "POSSIBLE"

Use each move to change each of the letters once.

"aaa"

"bab"

Returns: "POSSIBLE"

The following sequence of 4 moves does the job:
aaa -> baa -> bab -> aab -> bab

"aababbabaa"

"abbbbaabab"

Returns: "IMPOSSIBLE"

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

字符串水题！

249.23PT�Q�！�Q?/p>

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define min(x,y) (x
#define max(x,y) (x>y?x:y)
#define swap(t,x,y) (t=x,x=y,y=t)
#define clr(list) memset(list,0,sizeof(list))

using namespace std;

class EasyConversionMachine{
public:
    string isItPossible(string originalWord, string finalWord, int k)
    {
        int n=originalWord.size();
        int tot=0;
        for (int i=0;i<n;i++)
            if (originalWord[i]!=finalWord[i])
                tot++;
        if (tot<=k && (k-tot)%2==0) //��d��Q�这里刚刚开始搞反了�Q�testing WA了一��?br />            return "POSSIBLE";
        return "IMPOSSIBLE";
    }
};

wangs 2012-07-22 10:02 发表评论

TC SRM444 div2 250

wangs — Thu, 07 Jun 2012 16:39:00 GMT

�W�一�ơ准备做做TC�Q�发��C��么都不会(x��)�Q�SRM545开始了�Q�不能注册。只能做做SRM444 div2了，题目也是很久很久才看明白�Q�英语弱爆了�Q�C++��q��了啊�Q�！

题目大意很简单，��是一个贪心的�q�程吧，很简单的。不�l�说

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;

class FourBlocksEasy
{
public:
    int maxScore(vector <string> grid) //感觉�q�些参数传递其实挺��单的啊，
    {
        int i,j,tmp,num1=0,num2,len;
        int flag[2][55];
        len=grid[0].length();
        memset(flag,0,sizeof(flag));
        for(i=2;i<len-2;i++)
        {
            if(grid[0][i]=='1') flag[0][i]=1;
            if(grid[1][i]=='1') flag[1][i]=1;
        }
        for(i=2;i<len-2;i++)
        {
            if(flag[0][i]==0)
            {
                if((i+1<len-2)&& flag[0][i+1]==0&&flag[1][i]==0&&flag[1][i+1]==0)
                {
                    flag[0][i]=4;flag[0][i+1]=4;flag[1][i+1]=4;flag[1][i]=4;
                    num1+=4;
                }
            }
        }
        num2=2*(len-4);
        tmp=4*num1+num2-num1;
        return tmp;      //��q��做一个函数吧�Q?br />    }
}t1;
int main()
{
    vector<string> s1; //vector 的��用，什么意思啊�Q?br />    string s;
    getline(cin,s);
   s1.push_back(s); //擦。又是个什么函敎ͼ�看不懂�?br />    getline(cin,s);
    s1.push_back(s);
    int ans;
    ans=t1.maxScore(s1);
    printf("%d\n",ans);
    return 0;
}

贴的是jjjh的代码，哎，C++真是该学学哇�Q�连一个类都不�?x��)写�Q�啥也不�?x��)？�Q�？

以后要多做做TC!!!

oh ,shit..我是个煞�W�，居然自己testing都没有过��提交了�Q�还for 239.16。尼玛，坑爹啊！
不科学。。。�?br />
乖乖学C++吧�?br />

wangs 2012-06-08 00:39 发表评论

hdu 2093(�l�构体排�?水题哇哇�?

wangs — Thu, 07 Jun 2012 07:07:00 GMT

hdu 2093
�l�构体排�?
struct student
{
char name[20];
int AC,time;
}
首先按照AC�Q�time�Q�如果前两个都相同就按照name的字典序排序�?br />
�q�个��d��很简单，sscanf();�q�个函数�q�不�?x��)用�?br />
输出需要名字占10�Q�靠左，则用“%-10s”

直接贴代码：(x��)WA了一�ơ，中间调试输出的东西忘记删除了�?br />�ȝ��Q�字�W�串�c�题目还是要自己做做�Q�虽然有些很恶心�Q�恶心两�ơ就好了�Q�哈哈，

#include<stdio.h>
#include<string.h>
#include<math.h>
struct student
{
    char name[20];
    int AC,time;
} p[1005],temp,tt;

int tot;

int cmp(student a,student b)
{
    if (a.AC>b.AC)
        return 1;
    if (a.AC==b.AC&&a.time<b.time)
        return 1;
    int i=0;
    while (a.AC==b.AC&&a.time==b.time&&i<strlen(a.name)&&i<strlen(b.name))
    {
        if (a.name[i]<b.name[i])
            return 1;
        i++;
    }
    return 0;
}

int qsort(int l,int r)      //自己�q�是不习(f��n)惯条用stl里的sort()
{
    int i,j;
    i=l;
    j=r;
    temp=p[(i+j)/2];
    while (i<=j)
    {
        while (i<=j&&cmp(p[i],temp))    i++;
        while (i<=j&&cmp(temp,p[j]))    j--;
        if (i<=j)
        {
            tt=p[i];
            p[i]=p[j];
            p[j]=tt;
            i++;
            j--;
        }
    }
    if (l<j)    qsort(l,j);
    if (i<r)    qsort(i,r);
    return 0;
}

int main()
{
    int n,m,i,j,tmp;
    char ch;
    scanf("%d%d",&n,&m);
    tot=0;
    while (scanf("%s",&p[0].name)==1&&p[0].name[0]!='0')
    {
        tot++;
        p[tot]=p[0];
        p[tot].AC=0;
        p[tot].time=0;
        for (i=1; i<=n ; i++ )
        {
            scanf("%d",&tmp);
            if (tmp>0)
            {
                p[tot].AC++;
                p[tot].time+=tmp;
                ch=getchar();
                if (ch=='(')
                {
                    scanf("%d",&tmp);
                    p[tot].time+=m*tmp;
                    getchar();
                }
            }
        }
    }

    qsort(1,tot);

    for (i=1; i<=tot ; i++ )
    {

        printf("%-10s %2d %4d\n",p[i].name,p[i].AC,p[i].time);
    }
    return 0;
}

wangs 2012-06-07 15:07 发表评论