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acpeng — Tue, 13 Dec 2011 14:06:00 GMT

又刷一道水题，地址�Q?a >http://acm.fzu.edu.cn/problem.php?pid=2064

内容不复�q�ͼ�直接code

acpeng 2011-12-13 22:06 发表评论

HDU 2084 数塔

acpeng — Mon, 03 Oct 2011 04:16:00 GMT

题目链接�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=2084

�l�典入门�U�DP�Q�自底向上，层层递进。（所谓的记忆化搜索）(j��)

#include<stdio.h>
#define max(a,b) (a)>(b)?(a):(b)
int a[103][103];
int main()
{
    int t,n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            for(j=0;j<=i;j++)
                scanf("%d",&a[i][j]);
        for(i=n-1;i>0;i--)
            for(j=0;j<i;j++)
                a[i-1][j]+=max(a[i][j],a[i][j+1]);
        printf("%d\n",a[0][0]);
    }
    return 0;
}

acpeng 2011-10-03 12:16 发表评论

HDU 1160 ZOJ 1108 FatMouse's Speed

acpeng — Sun, 02 Oct 2011 19:24:00 GMT

题目链接�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=1160

Problem Description

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.

Input

Input contains data for a bunch of mice, one mouse per line, terminated by end of file.
The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.
Two mice may have the same weight, the same speed, or even the same weight and speed.

Output

Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that

W[m[1]] < W[m[2]] < ... < W[m[n]]

and

S[m[1]] > S[m[2]] > ... > S[m[n]]

In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.

Sample Input

Sample Output

4
4
5
9
7
解题�?/span>路：(x��)
首先�Ҏ(gu��)��weight的数值由���到大排序，然后再以speed为标准，求最长下降子序列�Q�思�\�q�是DP�Q�但是要标记子序列的下标
�ȝ��(ch��)多了�Q�我的做法：(x��)
定义MaxIncludeEnd数组�Q���o(h��)MaxIncludeEnd[i]表示以a[i]�l�尾的最长下降子序列的长度，那么
MaxIncludeEnd[i]=max{ MaxIncludeEnd[i] , a[j]>a[i] ? (MaxIncludeEnd[j]+1) �Q?1; }
其中j=0:i-1;MaxIncludeEnd[i]的初始��gؓ(f��)1,每一个i�l�束后用max和MaxIncludeEnd[i]比较�Q�更新max和下标max_flg�?/span>
因�ؓ(f��)要标��C��标，在计���MaxIncludeEnd[i]后，用flg[i]表示以a[i]�l�尾的最长下降子序列的前一个元素的下标�Q�这栯������完
MaxIncludeEnd[n]�q�更新完max和max_flg以后�Q���@环递推一�ơ就可以得到所有的元素下标�Q�即flg[max_flg],flg[ flg[max_flg] ],……
当然�q�里的下标是逆向的，要设�|�个新的数组来将光����{�?/span>
ps�Q�如果从后往前计���最长上升子序列�Q�那��׃��必如此麻�?ch��)�?/span>减少�I�间�?div style="border-bottom: #cccccc 1px solid; border-left: #cccccc 1px solid; padding-bottom: 4px; background-color: #eeeeee; padding-left: 4px; width: 98%; padding-right: 5px; font-size: 13px; word-break: break-all; border-top: #cccccc 1px solid; border-right: #cccccc 1px solid; padding-top: 4px">code
#include<cstdio>
#include<algorithm>
using namespace std;
struct mouse
{
    int weight;
    int speed;
    int num;
}mice[1002];
int MaxIncludeEnd[1002],flg[1002],rst[1002];
int cmp(struct mouse a,struct mouse b)
{
    if(a.weight<b.weight)return 1;
    return 0;
}
int main()
{
    int w,s,i=0,j,k,size,max,max_flg;
    while(scanf("%d%d",&w,&s)!=EOF)
    {
        mice[i].weight=w;
        mice[i].speed=s;
        mice[i].num=i+1;
        i++;
    }
    size=i;
    sort(mice,mice+size,cmp);
    max=max_flg=flg[0]=0;
    for(i=0;i<size;i++)
    {
        MaxIncludeEnd[i]=1;
        for(j=0;j<i;j++)
        {
            if(mice[i].speed<mice[j].speed && mice[i].weight>mice[j].weight && MaxIncludeEnd[j]+1>MaxIncludeEnd[i])
            {
                MaxIncludeEnd[i]=MaxIncludeEnd[j]+1;
                flg[i]=j;
            }
        }
        if(max<MaxIncludeEnd[i])
        {
            max=MaxIncludeEnd[i];
            max_flg=i;
        }
    }
    printf("%d\n",max);
    k=0;
    while(max--)
    {
        rst[k++]=max_flg;
        max_flg=flg[max_flg];
    }
    for(k--;k>=0;k--)
        printf("%d\n",mice[rst[k]].num);
    return 0;
}

acpeng 2011-10-03 03:24 发表评论

HDU 1087 Super Jumping

acpeng — Sun, 02 Oct 2011 17:28:00 GMT

题目链接�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=1087

Problem Description

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard�Q�棋盘）(j��)and some chessmen�Q�棋子）(j��), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.

Input

Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each case, print the maximum according to rules, and one line one case.

Sample Input

Sample Output

4
10
3













典型的入门��DP题，解题思�\�Q?br />定义MaxIncludeEnd数组�Q���o(h��)MaxIncludeEnd[i]表示以a[i]�l�尾的最大和�Q�那�?br />MaxIncludeEnd[i]=max{ MaxIncludeEnd[i] , a[j]其中j=0 to i-1;MaxIncludeEnd[i]的初始��gؓ(f��)a[i]�?br />#include<stdio.h>
int a[1003],MaxIncludeEnd[1003];
int main()
{
    int n,max,i,j;
    while(scanf("%d",&n),n)
    {
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            MaxIncludeEnd[i]=a[i];
        }
        max=MaxIncludeEnd[0];
        for(i=0;i<n;i++)
        {
            for(j=0;j<i;j++)
            {
                if(a[j]<a[i] && MaxIncludeEnd[j]+a[i]>MaxIncludeEnd[i])
                    MaxIncludeEnd[i]=MaxIncludeEnd[j]+a[i];
            }
            if(max<MaxIncludeEnd[i])max=MaxIncludeEnd[i];
        }
        printf("%d\n",max);
    }
    return 0;
}

acpeng 2011-10-03 01:28 发表评论

ZOJ1292 , HDU 1047 Integer Inquiry

acpeng — Mon, 05 Sep 2011 15:22:00 GMT

大数相加�Q�字�W�模�?br />题目链接�Q?br />ZOJ�Q?a >http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1292
HDU�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=1047

Description
One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.

``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)

Input
The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).

The final input line will contain a single zero on a line by itself.

Output
Your program should output the sum of the VeryLongIntegers given in the input.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

AC代码�Q?/p>

/*
Accepted 1047 0MS 188K 1256 B C++
*/
#include<stdio.h>
#include<string.h>
#define MaxSize 1000
void FunBigAdd(char *str1,char *str2);
int main()
{
    char IntStr[MaxSize]="\0",str[MaxSize]="\0";
    int T,i;
    scanf("%d",&T);
    while(T--)
    {
        memset(str,0,MaxSize);
        scanf("%s",IntStr);
        while(strcmp(IntStr,"0")!=0)
        {
            FunBigAdd(str,IntStr);
            memset(IntStr,0,MaxSize);
            scanf("%s",IntStr);
        }
        i=0;
        while(str[i]=='0' && str[i]!='\0')i++;
        if(str[i]=='\0')printf("0\n");
        else printf("%s\n",str+i);
        if(T!=0)printf("\n");
    }
    return 0;
}
void strRevg(char *str)
{
    int i;char temp;
    int lgth=(int)strlen(str)-1;
    for(i=0;i<lgth-i;i++)
    {
        temp=*(str+i);
        *(str+i)=*(str+lgth-i);
        *(str+lgth-i)=temp;
    }
}
void FunBigAdd(char *str1,char *str2)
{
    int i=0,dec=0,tmp;
    strRevg(str1);
    strRevg(str2);
    while(*(str1+i)!='\0' || *(str2+i)!='\0')
    {
        if(*(str1+i)=='\0')
        {
            *(str1+i)='0';
            *(str1+i+1)='\0';
        }
        if(*(str2+i)=='\0')
        {
            *(str2+i)='0';
            *(str2+i+1)='\0';
        }
        tmp=*(str1+i)-'0'+(*(str2+i)-'0')+dec;
        *(str1+i)=tmp%10+'0';
        dec=tmp/10;
        i++;
    }
    if(dec!=0) *(str1+(i++))=dec+'0';
    *(str1+i)='\0';
    strRevg(str1);
}

acpeng 2011-09-05 23:22 发表评论

acpeng — Mon, 05 Sep 2011 14:47:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=61
数据�l�构�Q�用C写的栈，不用STL��实复杂多了�Q�W(xu��)A了几�ơ~
Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be reached by moving backward and forward. In this problem, you are asked to implement this.

The following commands need to be supported:

BACK: Push the current page on the top of the forward stack. Pop the page from the top of the backward stack, making it the new current page. If the backward stack is empty, the command is ignored.

FORWARD: Push the current page on the top of the backward stack. Pop the page from the top of the forward stack, making it the new current page. If the forward stack is empty, the command is ignored.

VISIT : Push the current page on the top of the backward stack, and make the URL specified the new current page. The forward stack is emptied.

QUIT: Quit the browser.

Assume that the browser initially loads the web page at the URL http://www.acm.org/

This problem contains multiple test cases!

The output format consists of N output blocks. There is a blank line between output blocks.

Input

Input is a sequence of commands. The command keywords BACK, FORWARD, VISIT, and QUIT are all in uppercase. URLs have no whitespace and have at most 70 characters. You may assume that no problem instance requires more than 100 elements in each stack at any time. The end of input is indicated by the QUIT command.

Output

For each command other than QUIT, print the URL of the current page after the command is executed if the command is not ignored. Otherwise, print "Ignored". The output for each command should be printed on its own line. No output is produced for the QUIT command.

Sample Input

VISIT http://acm.ashland.edu/
VISIT http://acm.baylor.edu/acmicpc/
BACK
BACK
BACK
FORWARD
VISIT http://www.ibm.com/
BACK
BACK
FORWARD
FORWARD
FORWARD
QUIT

Sample Output

http://acm.ashland.edu/
http://acm.baylor.edu/acmicpc/
http://acm.ashland.edu/
http://www.acm.org/
Ignored
http://acm.ashland.edu/
http://www.ibm.com/
http://acm.ashland.edu/
http://www.acm.org/
http://acm.ashland.edu/
http://www.ibm.com/
Ignored

AC代码�Q?/p>

#include<stdio.h>
#include<string.h>
#include<malloc.h>
#define StackSize 105
typedef struct
{
    char elemt[StackSize][73];
    int top;
}Stack;
void InitStack(Stack *S)
{
      S->top=-1;
}
int ElemtNum(Stack *S)
{
    return (S->top)+1;
}
int IsFull(Stack *S)
{
    return(S->top==StackSize-1?1:0);
}
int PushStack(Stack *S,char *x)
{
    if(IsFull(S))
        return 0;
    S->top++;
    memset(S->elemt[S->top],'\0',sizeof(S->elemt[S->top]));
    strcpy(S->elemt[S->top],x);
    return 1;
}
int PopStack(Stack *S,char *x)
{
    if(ElemtNum(S)==0)
        return 0;
    memset(x,0,sizeof(x));
      strcpy(x,S->elemt[S->top]);
    memset(S->elemt[S->top],0,sizeof(S->elemt[S->top]));
    S->top--;
      return 1;
}
int GetTopElemt(Stack *S,char *x)
{
    if(ElemtNum(S)==0)
        return 0;
      memset(x,0,sizeof(x));
      strcpy(x,S->elemt[S->top]);
      return 1;
}
int main()
{
    Stack *sBack,*sForward;
    sBack=(Stack *)malloc(sizeof(Stack));
    sForward=(Stack *)malloc(sizeof(Stack));
    int T;char str[73]="\0",comd[8]="\0",tmp[73]="\0";
    scanf("%d",&T);
    while(T--)
    {
        InitStack(sBack); InitStack(sForward);
        PushStack(sBack,"http://www.acm.org/");
        while(1)
        {
            memset(comd,0,sizeof(comd));
            scanf("%s",comd);
            if(strcmp(comd,"BACK")==0)
            {
                if(ElemtNum(sBack)==1)
                {
                    printf("Ignored\n");
                    continue;
                }
                PopStack(sBack,tmp);
                PushStack(sForward,tmp);
                if(GetTopElemt(sBack,tmp)==1)
                    printf("%s\n",tmp);
            }
            else if(strcmp(comd,"FORWARD")==0)
            {
                if(ElemtNum(sForward)==0)
                {
                    printf("Ignored\n");
                    continue;
                }
                PopStack(sForward,tmp);
                PushStack(sBack,tmp);
                if(GetTopElemt(sBack,tmp)==1)
                    printf("%s\n",tmp);
            }
            else if(strcmp(comd,"VISIT")==0)
            {
                memset(str,0,sizeof(str));
                scanf("%s",str);
                printf("%s\n",str);
                PushStack(sBack,str);
                InitStack(sForward);
            }
            else if(strcmp(comd,"QUIT")==0)
                break;
        }
        if(T>0) printf("\n");
    }
    free(sBack);free(sForward);
    return 0;
}

acpeng 2011-09-05 22:47 发表评论

HDU 1862 EXCEL排序

acpeng — Mon, 05 Sep 2011 14:20:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1862
�l�构体排序，C++的algorithm�Q�写三个cmp卛_��
不过以这个数据的话，何必TimeLimit=5000ms~~

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef struct infor
{
    char stu_num[7];
    char name[9];
    int scor;
}stu;
stu data[100002];
int cmp1(stu a,stu b)
{
    if(strcmp(a.stu_num,b.stu_num)<0)return 1;
    return 0;
}
int cmp2(stu a,stu b)
{
    if(strcmp(a.name,b.name)<0)return 1;
    else if(strcmp(a.name,b.name)==0)
    {
        if(strcmp(a.stu_num,b.stu_num)<0)return 1;
    }
    return 0;
}
int cmp3(stu a,stu b)
{
    if(a.scor<b.scor)return 1;
    else if(a.scor==b.scor)
    {
        if(strcmp(a.stu_num,b.stu_num)<0)return 1;
    }
    return 0;
}
int main()
{
    int n,c,i,test=1;
    while(scanf("%d%d",&n,&c),n)
    {
        for(i=0;i<n;i++)
            scanf("%s%s%d",data[i].stu_num,data[i].name,&data[i].scor);
        if(c==1)sort(data,data+n,cmp1);
        else if(c==2)sort(data,data+n,cmp2);
        else sort(data,data+n,cmp3);
        printf("Case %d:\n",test++);
        for(i=0;i<n;i++)
            printf("%s %s %d\n",data[i].stu_num,data[i].name,data[i].scor);
    }
    return 0;
}

acpeng 2011-09-05 22:20 发表评论

POJ 2273 An Excel-lent Problem

acpeng — Mon, 05 Sep 2011 13:53:00 GMT

把整数�{化�ؓ(f��)excel对应的列�?a >http://poj.org/problem?id=2273
Description

A certain spreadsheet program labels the columns of a spreadsheet using letters. Column 1 is labeled as ``A", column 2 as ``B", ..., column 26 as ``Z". When the number of columns is greater than 26, another letter is used. For example, column 27 is ``AA", column 28 is ``AB" and column 52 is ``AZ". It follows that column 53 would be ``BA" and so on. Similarly, when column ``ZZ" is reached, the next column would be ``AAA", then ``AAB" and so on.

The rows in the spreadsheet are labeled using the row number. Rows start at 1.

The designation for a particular cell within the spreadsheet is created by combining the column label with the row label. For example, the upper-left most cell would be ``A1". The cell at column 55 row 23 would be ``BC23".

You will write a program that converts numeric row and column values into the spreadsheet designation.

Input

Input consists of lines of the form: RnCm. n represents the row number [1,300000000] and m represents the column number, 1<=m<=300000000. The values n and m define a single cell on the spreadsheet. Input terminates with the line: R0C0 (that is, n and m are 0). There will be no leading zeroes or extra spaces in the input.

Output

For each line of input (except the terminating line), you will print out the spreadsheet designation for the specified cell as described above.

Sample Input

R1C1
R3C1
R1C3
R299999999C26
R52C52
R53C17576
R53C17602
R0C0

Sample Output

A1
A3
C1
Z299999999
AZ52
YYZ53
YZZ53

Source

Greater New York 2004

�q�制转化问题�Q�把列数�Ҏ(gu��)��26�q�制输出�Q�分别用A,B...Z代替1,2,3...�q�里base[0]='Z'�Q�表�C�n�?6余数�?时的取��|��但进位问题需要注意，分余数是否�ؓ(f��)0两种情况分析�?br />代码�Q?

#include<stdio.h>
#define R 26
char base[30]="ZABCDEFGHIJKLMNOPQRSTUVWXYZ";
void fun(int N)
{
    int i=0,j; char str[10];
    while(N>R)
    {
        str[i++]=base[N%R];
        N=N%R==0?N/R-1:N/R;
    }
    str[i]=base[N];
    for(j=i;j>=0;j--)
        printf("%c",str[j]);
}
int main()
{
    int m,n;
    while(scanf("%*c%d%*c%d",&m,&n)!=EOF)
    {
        getchar();
        if(m==0 && n==0)break;
        fun(n);
        printf("%d\n",m);
    }
    return 0;
}

acpeng 2011-09-05 21:53 发表评论

HDU 1166敌兵布阵

acpeng — Thu, 01 Sep 2011 09:14:00 GMT

题目链接�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=1166
初学�U�段�?w��i)，本想些个树(w��i)，写了一半实在是觉得没必要，此题直接用数�l�模拟，果断AC�Q?Y~~~

#include<stdio.h>
#include<string.h>
int data[50003];
int main()
{
    int T,N,i,cont=1;
    int cd_1,cd_2,sum;
    char comd[7]="\0";
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        printf("Case %d:\n",cont);
        cont++;
        for(i=0;i<N;i++)
            scanf("%d",&data[i]);
        scanf("%s",comd);
        while(strcmp(comd,"End")!=0)
        {
            scanf("%d%d",&cd_1,&cd_2);
            if(comd[0]=='A')
                data[cd_1-1]+=cd_2;
            else if(comd[0]=='S')
                data[cd_1-1]-=cd_2;
            else if(comd[0]=='Q')
            {
                sum=0;
                for(i=cd_1-1;i<cd_2;i++)
                    sum+=data[i];
                printf("%d\n",sum);
            }
            memset(comd,0,sizeof(comd));
            scanf("%s",comd);
        }
    }
    return 0;
}

acpeng 2011-09-01 17:14 发表评论

acpeng — Fri, 26 Aug 2011 10:04:00 GMT

题目地址�Q?a >http://acm.hdu.edu.cn/showproblem.php?pid=1022

Problem Description

As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.

Input

The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.

Output

The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.

Sample Input

3 123 321
3 123 312

Sample Output

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH
下面是代码，�W�一�ơ��用C++的STL模板�Q�用的很不熟�l�，不过没想�?Y���p��了~~~YY。。�?div style="border-bottom: #cccccc 1px solid; border-left: #cccccc 1px solid; padding-bottom: 4px; background-color: #eeeeee; padding-left: 4px; width: 98%; padding-right: 5px; font-size: 13px; word-break: break-all; border-top: #cccccc 1px solid; border-right: #cccccc 1px solid; padding-top: 4px">#include<cstdio>
#include<stack>
using namespace std;
int main()
{
    int i,j,k,n,ok,flg[22];
    char in[10],out[10];
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s%s",in,out);
        stack<int> s;
        i=j=k=0;ok=1;
        while(j<n)
        {
            if(in[i]==out[j])
            {
                flg[k++]=0;flg[k++]=1;i++;j++;
            }
            else if(!s.empty() && s.top()==(out[j]-'0'))
            {
                flg[k++]=1;s.pop();j++;
            }
            else if(i<n)
            {
                s.push(in[i++]-'0');flg[k++]=0;
            }
            else 
            {
                ok=0;break;
            }
        }
        if(ok==0)printf("No.\n");
        else
        {
            printf("Yes.\n");
            for(i=0;i<k;i++)
            {
                if(flg[i]==0)printf("in\n");
                else printf("out\n");
            }
        }
        printf("FINISH\n");
    }
    return 0;
}

acpeng 2011-08-26 18:04 发表评论