poj 1306 Combinations

#include<iostream>//递归
usingnamespace std;
__int64 ans[200][200];
__int64 f(int n,int m)
{
    if(ans[n][m]>0)
        return ans[n][m];
    if(n==m||m==0)
        return1;
    ans[n][m]=f(n-1,m)+f(n-1,m-1);        //C(n,m)=C(n-1,m)+C(n-1,m-1)
return ans[n][m];
}
int main()
{
    int n,m;
    while(cin>>n>>m&&n)
    {
        memset(ans,0,sizeof(ans));
        printf("%d things taken %d at a time is %I64d exactly.\n",n,m,f(n,m));
    }
    return0;
}

　　

 

 

#include<iostream>
usingnamespace std;
int main()
{
    double n,m,i,j,res;
    while(cin>>n>>m&&n)
    {
        res=1;
        double r1=1,r2=1;
        for(i=n-m+1,j=1;i<=n;i++,j++)    // res = (n-m+1)*(n-m+2)*...*n  / 1*2*...*m
        {
            r1*=i;
            r2*=j;    
        }
        res=r1/r2;
        printf("%.0f things taken %.0f at a time is %.0f exactly.\n",n,m,res);
    }
    return0;
}



#include<iostream>
usingnamespace std;
int main()
{
    double n,m,i,j,res;
    while(cin>>n>>m&&n)
    {
        res=1;
        for(i=n-m+1,j=1;i<=n;i++,j++)    // res = (n-m+1)*(n-m+2)*...*n  / 1*2*...*m
        {
            res*=i/j;    
        }
        printf("%.0f things taken %.0f at a time is %.0f exactly.\n",n,m,res);
    }
    return0;
}

　

 

　

// 题意:求组合数C(n,m) = n! / (n-m)!m!
#include <iostream>
#include <string>
using namespace std;

int compare(string str1, string str2)
{
    while(!str1.empty()&&str1[0]=='0')
    {
        str1.erase(0,1);
    }
    while(!str2.empty()&&str2[0]=='0')
    {
        str2.erase(0,1);
    }
    if(str1.size() > str2.size()) //长度长的整数大于长度小的整数
        return 1;
    else if(str1.size() < str2.size())
        return -1;
    else
        return str1.compare(str2); //若长度相等，从头到尾按位比较，compare函数：相等返回0，大于返回1，小于返回－1
}
int to_int(char ch)
{
    return ch-48;
}
char to_char(int i)
{
    return (char)(i+48);
}
string to_str(int n)
{
    string str;
    while(n)
    {
        str.insert(str.begin(),n%10+48);
        n=n/10;
    }
    return str;
}
void dipose_head(string& str,int sign)    //str为引用类型
{
    str.erase(0, str.find_first_not_of('0'));    //去除结果中的前导0
    if(str.empty()) str = "0";
    if((sign == -1) && (str[0] != '0'))        //处理符号位 ,若str[0] == '0'说明结果是0,就不必要在前面加上-
        str = '-' + str;
}

//高精度加法
string ADD_INT(string str1, string str2)        //计算str1 + str2的值,处理成 str1 ,str2 >= 0
{
    if(str1.empty())    str1="0";
    if(str2.empty())    str2="0";
    string MINUS_INT(string str1, string str2);        //要事先声明MINUS_INT
    string str;
    if(str1[0] == '-')
    {
        if(str2[0] == '-') 
        {
            str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));
            if(str[0] != '0')
                str = "-" + str;
        }
        else 
            str = MINUS_INT(str2, str1.erase(0, 1));
    }
    else 
    {
        if(str2[0] == '-')
            str = MINUS_INT(str1, str2.erase(0, 1));

        else    //str1 ,str2 >= 0的情况
        {
            string::size_type l1, l2;
            int i;
            l1 = str1.size(); l2 = str2.size();
            if(l1 < l2)        //把两个整数对齐，短整数前面加0补齐
                for(i = 1; i <= l2 - l1; i++)
                    str1 = '0' + str1;
            else
                for(i = 1; i <= l1 - l2; i++)
                    str2 = '0' + str2;
            int int1 = 0, int2 = 0;        //int2 记录进位
            for(i = str1.size() - 1; i >= 0; i--)
            { 
                int1 = ( str1[i]+str2[i] -  96  + int2 ) % 10;  
                int2 = ( str1[i]+str2[i] -  96  + int2 ) / 10; 
                str = to_char(int1) + str;
            }
            if(int2 != 0) str = to_char(int2) + str;    //处理最后的进位
        }
    }
    return str;
}

//高精度减法
string MINUS_INT(string str1, string str2)        //计算str1 - str2的值,先处理成 str1 ,str2 >= 0,再处理成 str1 > str2 >= 0
{
    if(str1.empty())    str1="0";
    if(str2.empty())    str2="0";
    int sign = 1;       //sign 为符号位
    string str;
    if(str2[0] == '-')
    {
        if(str1[0]!='-')
            str = ADD_INT(str1, str2.erase(0, 1));
        else 
            str=MINUS_INT(str2.erase(0,1),str1.erase(0,1));
    }
    else 
    {
        if(str1[0]=='-')
        {
            str = ADD_INT(str1.erase(0, 1), str2);
            if(str[0] != '0')
                str = "-" + str;
        }
        else    //str1 ,str2 >= 0的情况
        {
            int res = compare(str1, str2);
            if(res == 0) return "0";      //二者相等
            if(res < 0) 
            {
                sign = -1;
                str1.swap(str2);
            }
            string::size_type temp_int;    
            temp_int = str1.size() - str2.size();
            int cf=0;        //cf保存借位
            for(int i = str2.size() - 1; i >= 0; i--) 
            {
                if(str1[i + temp_int] - cf < str2[i]) 
                {
                    str = to_char(str1[i + temp_int] - cf - str2[i] + 10) + str;
                    cf=1;
                }
                else
                {
                    str =to_char(str1[i + temp_int] - cf - str2[i]) + str;
                    cf=0;
                }
            }
            for(int i=temp_int-1;i>=0;--i)
            {
                if(str1[i]-cf>='0')        //看是否够借位减
                {
                    str=char(str1[i]-cf)+str;
                    cf=0;
                }
                else
                {
                    str=char(str1[i]-cf+10)+str;
                    cf=1;
                }
            }
            dipose_head(str,sign);
        } 
    }
    return str;
}

//高精度乘法
string MULTIPLY_INT(string str1, string str2)
{
    if(str1.empty())    str1="0";
    if(str2.empty())    str2="0";
    int sign = 1; //sign 为符号位
    if(str1[0] == '-') 
    {
        sign *= -1;
        str1 = str1.erase(0, 1);
    }
    if(str2[0] == '-') 
    {
        sign *= -1;
        str2 = str2.erase(0, 1);
    }
    if(str1=="0"||str2=="0")    return "0";

    int i, j;    string str;
    string::size_type len1, len2;
    len1 = str1.size(); len2 = str2.size();

    for(i = len2 - 1; i >= 0; i --)       //实现手工乘法
    { 
        string temp_str;
        int int1 = 0, int2 = 0, int3 = to_int(str2[i]);
        if(int3 != 0) 
        {
            for(j = 1; j <= len2-1 - i; j++)
                temp_str = '0' + temp_str;
            for(j = len1 - 1; j >= 0; j--) 
            {
                int1 = (int3 * to_int(str1[j]) + int2) % 10;
                int2 = (int3 * to_int(str1[j]) + int2) / 10;
                temp_str = to_char(int1) + temp_str;
            }
            if(int2 != 0) temp_str = to_char(int2) + temp_str;
        }
        str = ADD_INT(str, temp_str);
    }
    if((sign == -1) && (str[0] != '0'))        
        str = '-' + str;
    return str;
}

//高精度除法
string DIVIDE_INT(string str1, string str2, int flag)        //flag = 1时,返回商; flag = 0时,返回余数
{
    if(str1.empty())    str1="0";
    if(str2.empty())    str2="0";
    string quotient, residue; //定义商和余数，sign1,sign2判断商和余数的符号
    int sign1 = 1, sign2 = 1;
    if(str2 == "0")        //判断除数是否为0 
        return "ERROR!";
    if(str1 == "0")        //判断被除数是否为0
        return "0";

    if(str1[0] == '-')
    {
        str1 = str1.erase(0, 1);
        sign1 *= -1;
        sign2 = -1;        //余数的符号由被除数的符号决定
    }
    if(str2[0] == '-') 
    {
        str2 = str2.erase(0, 1);
        sign1 *= -1;
    }
    int res = compare(str1, str2);
    if(res < 0) 
    {
        quotient = "0";
        residue = str1;
    }
    else if(res == 0) 
    {
        quotient = "1";
        residue = "0";
    }
    else    //str1>str2
    {
        string::size_type len1, len2;
        len1 = str1.size(); len2 = str2.size();
        string temp_str;
        temp_str.append(str1, 0, len2 - 1);    //添加从str1[0]到str1[len2 - 2]

        for(int i = len2 - 1; i < len1; i++)         //模拟手工除法
        {
            temp_str = temp_str + str1[i];

            for(char ch = '9'; ch >= '0'; ch --)  //试商
            {    
                string str;
                str = str + ch;
                if(compare(MULTIPLY_INT(str2, str), temp_str) <= 0) 
                    //在compare增添的两个while()语句在这里发挥作用，倘若缺失,比如计算10/1,第二次的temp_str=00,按照compare函数的定义，00就会比1大了
                {

                    quotient = quotient + ch;
                    temp_str = MINUS_INT(temp_str, MULTIPLY_INT(str2, str));
                    break;
                }
            }
        }
        residue = temp_str;        //residue就不用判断前面是否带0，因为temp_str = MINUS_INT（...），MINUS_INT把0处理了

        if(quotient[0]=='0')
            quotient.erase(0,1);        //商前面最多只有一个0
    }

    if((sign1 == -1) && (quotient!="0"))        quotient = "-" + quotient;

    if((sign2 == -1) && (residue != "0"))        residue = "-" + residue;

    if(flag == 1) 
        return quotient;
    else 
        return residue;
}

string a_multiply_b(int a,int b)    //a*...*b
{
    string str=to_str(a),tmp;
    for(int i=a+1;i<=b;i++)
    {
        tmp=to_str(i);
        str=MULTIPLY_INT(str,tmp);
    }
    return str;
}

int main()
{
    int n,m;string m1,m2;
    while(scanf("%d%d",&n,&m)&&n)                                    
    {
        if(n==m)
            printf("%d things taken %d at a time is 1 exactly.\n",n,m);
        else
        {
            if(n-m<m)
            {
                m1=a_multiply_b(m+1,n);
                m2=a_multiply_b(1,n-m);
            }
            else
            {
                m1=a_multiply_b(n-m+1,n);
                m2=a_multiply_b(1,m);
            }
            cout<<n<<" things taken "<<m<<" at a time is "<<DIVIDE_INT(m1,m2,1)<<" exactly.\n";    
        }
    }
    return 0;
}