``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：

/**//*
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);
}

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 <url>: 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 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.

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

1

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：

#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;
}

把整数转化为excel对应的列值http://poj.org/problem?id=2273
Description

Input

Output

Sample Input

R1C1
R3C1
R1C3
R299999999C26
R52C52
R53C17576
R53C17602
R0C0

Sample Output

A1
A3
C1
Z299999999
AZ52
YYZ53
YZZ53

Source

Greater New York 2004

进制转化问题，把列数改成26进制输出，分别用A,B...Z代替1,2,3...这里base[0]='Z'，表示n与26余数为0时的取值，但进位问题需要注意，分余数是否为0两种情况分析。
代码：

#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;
}

100

Sample Output

354224848179261915075

CAPILLARY
MARSUPIAL

There is only one common letter (A). Better is the following overlay:

CAPILLARY
     MARSUPIAL

with two common letters (A and R), but the best is:

   CAPILLARY
MARSUPIAL

Which has three common letters (P, I and L).

The approximation measure appx(word1, word2) for two words is given by:

     common letters * 2
-----------------------------
length(word1) + length(word2)

Thus, for this example, appx(CAPILLARY, MARSUPIAL) = 6 / (9 + 9) = 1/3. Obviously, for any word W appx(W, W) = 1, which is a nice property, while words with no common letters have an appx value of 0.
Input:
The input for your program will be a series of words, two per line, until the end-of-file flag of -1.
Using the above technique, you are to calculate appx() for the pair of words on the line and print the result. For example:

CAR CART
TURKEY CHICKEN
MONEY POVERTY
ROUGH PESKY
A A
-1

The words will all be uppercase.
Output:
Print the value for appx() for each pair as a reduced fraction, like this:

appx(CAR,CART) = 6/7
appx(TURKEY,CHICKEN) = 4/13
appx(MONEY,POVERTY) = 1/3
appx(ROUGH,PESKY) = 0
appx(A,A) = 1

Fractions reducing to zero or one should have no denominator

代码：

若一个整数Y的个位数为X，将X移到最高位得到的恰好是Y的X倍，称Y为一个X倍数。例如，0是一个0倍数，1是一个1倍数。

Input

无

Output

请依次输出最小的0倍数、1倍数、2倍数、……9倍数。每个X倍数输出一行。

Sample Input

无

Sample Output

0

1

……

……

……

Hint

即使是最小的X倍数也可能是一个非常大的整数，甚至远远超出int所能表达的范围。
 
对于X倍数，设定一个数组（事实上应该是字符串，节省空间），第0位赋予初始值k，k=2,3,……9,依次与k相乘得到上一位，即可。注意循环跳出的条件。
代码:

Roman numerals are based on seven symbols: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500 and M = 1000.

Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1,000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I "1", II "2", III "3", V "5", VI "6", VII "7", etc., and the same for other bases: X "10", XX "20", XXX "30", L "50", LXXX "80"; CC "200", DCC "700", etc. At the fourth iteration, a subtractive principle is employed, with the base placed before the higher base: IV for "4", IX for "9", XL for "40", XC for "90", CD for "400", CM for "900".

The basic multiples of Roman numerals thus follow a pattern:

A practical way to write a Roman number is to consider the modern Arabic numeral system, and separately convert the thousands, hundreds, tens, and ones as given in the chart above. So, for instance, 1234 may be thought of as "one thousand and two hundreds and three tens and four", obtaining M (one thousand) + CC (two hundreds) + XXX (thirty) + IV (four), for MCCXXXIV. Thus eleven is XI (ten and one), 29 is XXIX (twenty and nine), and 2011 is MMXI (two thousand and ten and one). Note that the subtractive principle is not extended beyond the chart: for example, IL is not used for 49, rather this should be written as forty (XL) and nine (IX), or XLIX.

Given a list of numbers, you are to rearrange them so that if we write them as Roman numbers, they are in lexicographical order.

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

Each test case starts with an integer 1 ≤ n ≤ 10000. Then n numbers 0 < ai < 4000.

Output

For each test case, output the n numbers in specified order.

Sample Input

3
3
1 2 3
7
1 5 10 50 100 500 1000
11
4 5 6 7 8 9 10 11 12 13 14

Sample Output

1 2 3
100 500 1 50 1000 5 10
4 9 5 6 7 8 10 11 12 13 14

代码：

 1#include<stdio.h>
 2#include<string.h>
 3#include<algorithm>
 4using namespace std;
 5char strHundreds[10][5]={"C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
 6char strTens[10][5]={"X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
 7char strOnes[10][5]={"I","II","III","IV","V","VI","VII","VIII","IX"};
 8char strThousands[4][4]={"M","MM","MMM"};
 9typedef struct node
10{
11    char str[20];
12    int flg;
13}NODE;
14NODE data[10005];
15int num[10005]={0};
16int cmp(NODE b, NODE c)
17{
18    if(strcmp(b.str,c.str)<0)
19        return 1;
20    else
21        return 0;
22}
23int main()
24{
25    int i,T,n,num_temp_a;
26    char tmp[20]="\0";
27    scanf("%d",&T);
28    while(T--)
29    {
30        scanf("%d",&n);
31        for(i=0;i<n;i++)
32        {
33            scanf("%d",&num[i]);
34            num_temp_a=num[i];
35            if(num_temp_a/1000!=0)
36            {
37                strcat(tmp,strThousands[num_temp_a/1000-1]);
38                num_temp_a=num_temp_a%1000;
39            }
40            if(num_temp_a/100!=0)
41            {
42                strcat(tmp,strHundreds[num_temp_a/100-1]);
43                num_temp_a=num_temp_a%100;
44            }
45            if(num_temp_a/10!=0)
46            {
47                strcat(tmp,strTens[num_temp_a/10-1]);
48                num_temp_a=num_temp_a%10;
49            }
50            if(num_temp_a!=0)
51            {
52                strcat(tmp,strOnes[num_temp_a-1]);
53            }
54            strcpy(data[i].str,tmp);
55            data[i].flg=i;
56            memset(tmp,0,sizeof(tmp));
57        }
58        sort(data,data+n,cmp);
59        printf("%d",num[data[0].flg]);
60        for(i=1;i<n;i++)
61            printf(" %d",num[data[i].flg]);
62        printf("\n");
63    }
64    return 0;
65}
66