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Problem Description
We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
“Oh, God! How terrible! ”

Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds– 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

Input
Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.

Output
Output the minimum positive value that one cannot pay with given coins, one line for one case.

Sample Input
1 1 3
0 0 0

Sample Output
4

参考代码：

#include
#include
#include
int c1[1000],c2[10000];
int main()
{
	int n1,n2,n5;
	while(scanf("%d%d%d",&n1,&n2,&n5)==3,n1+n2+n5)
	{
		int n,i,j,k;
		n=n1+n2*2+n5*5;
		for(i=0;i<=n;i++)//init c1 c2
		{
			c1[i]=0;c2[i]=0;
		}
		for(i=0;i<=n1;i++)//只有 1
		{
			c1[i]=1;
		}
		for(i=0;i<=n1;i++)//存在1和2
		{
			for(j=0;j<=n2*2;j+=2)
			{
				c2[i+j]+=c1[i];
			}
		}
		for(k=0;k<=n1+n2*2;k++)
		{
			c1[k]=c2[k];c2[k]=0;
		}
		for(j=0;j<=n1+n2*2;j++)//存在1、2、5
		{
			for(k=0;k<=n5*5;k+=5)
			{
				c2[k+j]+=c1[j];
			}
		}
		for(i=0;i<=5*n5+2*n2+n1;i++)
		{
			c1[i]=c2[i];c2[i]=0;
		}
		bool flag=0;
		for(i=0;i<=n;i++)
		{
			if(c1[i]==0)
			{
				printf("%d\n",i);
				flag=1;
				break;
			}
		}
			if(!flag)
				printf("%d\n",n+1);
		}
		return 0;
	}

Problem Description
Reverse Polish notation (RPN) is a method for representing expressions in which the operator symbol is placed after the arguments being operated on.
Polish notation, in which the operator comes before the operands, was invented in the 1920s by the Polish mathematician Jan Lucasiewicz.
In the late 1950s, Australian philosopher and computer scientist Charles L. Hamblin suggested placing the operator after the operands and hence created reverse polish notation.

RPN has the property that brackets are not required to represent the order of evaluation or grouping of the terms.
RPN expressions are simply evaluated from left to right and this greatly simplifies the computation of the expression within computer programs.
As an example, the arithmetic expression (3+4)*5 can be expressed in RPN as 3 4 + 5 *.

Reverse Polish notation, also known as postfix notation, contrasts with the infix notation of standard arithmetic expressions in which the operator symbol appears between the operands. So Polish notation just as prefix notation.

Now, give you a string of standard arithmetic expressions, please tell me the Polish notation and the value of expressions.

Input
There’re have multi-case. Every case put in one line, the expressions just contain some positive integers(all less than 100, the number of integers less than 20), bi-operand operators(only have 3 kinds : +,-,*) and some brackets'(‘,’)’.
you can assume the expressions was valid.

Output
Each case output the Polish notation in first line, and the result of expressions was output in second line.
all of the answers are no any spaces and blank line.the answer will be not exceed the 64-signed integer.

Sample Input
1+2-3*(4-5)
1+2*(3-4)-5*6

Sample Output
Case 1:
– + 1 2 * 3 – 4 5
6
Case 2:
– + 1 * 2 – 3 4 * 5 6
-31

题目考察表达式的转换将中缀表达式转换成前置表达式（波兰式），并计算出结果。可以直接套模板……需要注意表达式中空格的位置；字符串需要用__int64存贮。

关于前置表达式和后置表达式的转换方法可以参考这篇博文：波兰式、逆波兰式与表达式求值

参考代码：

#include 
#include 
#include 
using namespace std;

char stack[500];                    
int top;                        //栈顶指针       
char output[500], input[500];        
int outLen;

int priority(char op)           //定义运算符优先级     
{
    if (op=='+' || op=='-')
        return 1;
    if (op=='*' || op=='/')
        return 2;
    else
        return 0;
}

bool isOperator(char op)                
{
    return (op=='+' || op=='-' || op=='*' || op=='/');
}

void Polish(char *s,int len)            
{
    memset(output,'\0',sizeof output);    
    outLen = 0;
    for (int i=len-1; i >= 0; --i)        
    {
        if (isdigit(s[i]))                
        {
            output[outLen++] = s[i];    
            while (i-1 >= 0 && isdigit(s[i-1]))
            {
                output[outLen++] = s[i-1];
                --i;
            }
            output[outLen++] = ' ';        
        }
        if (s[i]==')')                    
        {
            ++top;
            stack[top] = s[i];
        }
        while (isOperator(s[i]))        
        {                                                
            if (top==0 || stack[top]==')' || priority(s[i]) >= priority(stack[top])) 
            {
                ++top;
                stack[top] = s[i];
                break;
            }
            else
            {
                output[outLen++] = stack[top];
                output[outLen++] = ' ';
                --top;
            }
        }
        if (s[i]=='(')                    
        {
            while (stack[top]!=')')
            {
                output[outLen++] = stack[top];
                output[outLen++] = ' ';
                --top;
            }
            --top;    
        }
    }
    while (top!=0)                        
    {
        output[outLen++] = stack[top];
        output[outLen++] = ' ';
        --top;
    }
}

char DstBuf[200];
char* OP(char* op1,char* op2,char op)
{
    __int64 res = 0;
    if (op=='+')
        res = _atoi64(op1) + _atoi64(op2);
    else if (op=='-')
        res = _atoi64(op1) - _atoi64(op2);
    else if (op=='*')
        res = _atoi64(op1) * _atoi64(op2);
    else if (op=='/')
        res = _atoi64(op1) / _atoi64(op2);
    _i64toa(res,DstBuf,10);
    return DstBuf;
}

char cSt1[200][80], cSt2[200][80];
__int64 calc(char *s)                
{
    int top1=0, top2=0, i;
    for (i=0; s[i]; ++i)
    {
        if (s[i] && s[i] != ' ')
        {
            ++top1;
            sscanf(s+i,"%s",cSt1[top1]);        
            while (s[i] && s[i] != ' ')
                ++i;
        }
    }
    while (top1 != 0)    
    {
        if (!isdigit(cSt1[top1][0]))    
        {
            OP(cSt2[top2], cSt2[top2-1], cSt1[top1][0]);
            memcpy(cSt2[top2-1],DstBuf,sizeof DstBuf);
            --top2;                    
            --top1;                        
        }
        else
        {
            ++top2;                        
            memcpy(cSt2[top2],cSt1[top1],sizeof cSt1[top1]);
            --top1;                        
        }
    }
    return _atoi64(cSt2[1]);
}
int main()
{
    int T = 1;
    while (gets(input))
    {
        Polish(input, strlen(input));
        reverse(output,output+outLen-1);
        output[outLen-1] = '\0';
        printf("Case %d:\n%s\n",T++,output);
        printf("%I64d\n",calc(output));//注意要__int64
    }
    return 0;
}

Problem Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

Input
One N in one line, process to the end of file.

Output
For each N, output N! in one line.

Sample Input
1
2
3

Sample Output
1
2
6

要求计算10000以内的阶乘,典型的大数问题.这里用的是 五位一存的方法. 关于X位一存的解释可以看这篇文章. 注意特殊情况(0,1)下的处理.

#include 
#include 
#define MOD 100000
int main()
{
    int a[8000];    
    int i,k,n,t,la;
    while(~scanf("%d",&n))
    {
        if(n==0||n==1) {printf("1\n");continue;}

        a[0] = 1; 
        la = 1;
        t = 0;
        for(k=2; k<=n; k++)
        {
            for(i=0; i 0)   
            {    
                a[la++] = t;
                t = 0; 
            } 
        }
        printf("%d",a[la-1]);
        for(i=la-2; i>=0; i--)
            printf("%05d",a[i]);
        printf("\n");
    }
    return 0;
}

Problem Description
《Shui Hu Zhuan》，also 《Water Margin》was written by Shi Nai’an — an writer of Yuan and Ming dynasty. 《Shui Hu Zhuan》is one of the Four Great Classical Novels of Chinese literature. It tells a story about 108 outlaws. They came from different backgrounds (including scholars, fishermen, imperial drill instructors etc.), and all of them eventually came to occupy Mout Liang(or Liangshan Marsh) and elected Song Jiang as their leader.

In order to encourage his military officers, Song Jiang always made a rank list after every battle. In the rank list, all 108 outlaws were ranked by the number of enemies he/she killed in the battle. The more enemies one killed, one’s rank is higher. If two outlaws killed the same number of enemies, the one whose name is smaller in alphabet order had higher rank. Now please help Song Jiang to make the rank list and answer some queries based on the rank list.

Input
There are no more than 20 test cases.

For each test case:

The first line is an integer N (0<n&<200), indicating that there are N outlaws.

Then N lines follow. Each line contains a string S and an integer K(0<k<300), meaning an outlaw’s name and the number of enemies he/she had killed. A name consists only letters, and its length is between 1 and 50(inclusive). Every name is unique.

The next line is an integer M (0<m<200) ,indicating that there are M queries.

Then M queries follow. Each query is a line containing an outlaw’s name.
The input ends with n = 0

Output
For each test case, print the rank list first. For this part in the output ,each line contains an outlaw’s name and the number of enemies he killed.

Then, for each name in the query of the input, print the outlaw’s rank. Each outlaw had a major rank and a minor rank. One’s major rank is one plus the number of outlaws who killed more enemies than him/her did.One’s minor rank is one plus the number of outlaws who killed the same number of enemies as he/she did but whose name is smaller in alphabet order than his/hers. For each query, if the minor rank is 1, then print the major rank only. Or else Print the major rank, blank , and then the minor rank. It’s guaranteed that each query has an answer for it.

Sample Input
5
WuSong 12
LuZhishen 12
SongJiang 13
LuJunyi 1
HuaRong 15
5
WuSong
LuJunyi
LuZhishen
HuaRong
SongJiang
0

Sample Output
HuaRong 15
SongJiang 13
LuZhishen 12
WuSong 12
LuJunyi 1
3 2
5
3
1
2

简单排序，直接上代码：

#include
#include
#include