Spoj 2319 数位统计（0,1, 2^k1 这些数分成M份）

将0,1, 2^k-1 这些数分成M份，每份的值为该份的1的数字之和 ，求使最大的那一份最小      K <=100 ,M<=100

最大值最小化，二分的经典应用

思路：二分这个最大值X，然后看能不能组成M份 ，然后再调整

Cal(x)：求出0-x这些数字当中1出现的次数。

        其实实际分析一下就很简单，把x化作是二进制的数，那么从高位往低位走，遇见一个1,假使这个位置在i位置，那么答案ans=ans+F[i-1]+j*2^(i-1)+1,其中j表示的是到现在位置所遇见的1的个数，F[i]表示的是0-2^i-1中所出现的1的个数，然后再加上之前出现的1的个数乘以该位出现1所经历的次数，再加上这个位置上自己出现的1的个数1，那么就是0-x出现1的次数。

Get(t):从高位往下走，看该位置是否可以添加一个1，跟上面类似。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

const int MAXN = 110;
int const N = 120;

struct bign
{
    int len, s[MAXN];
    bign ()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign (int num)
    {
        *this = num;
    }
    bign (const char *num)
    {
        *this = num;
    }
    bign operator = (const int num)
    {
        char s[MAXN];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num)
    {
        for(int i = 0; num[i] == '0'; num++);//去前导0
        len = strlen(num);
        for(int i = 0; i < len; i++)
            s[i] = num[len-i-1] - '0';
        return *this;
    }
    bign operator + (const bign &b) const //+
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    bign operator += (const bign &b)
    {
        *this = *this + b;
        return *this;
    }
    void clean()
    {
        while(len > 1 && !s[len-1]) len--;
    }
    bign operator * (const bign &b) //*
    {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
            for(int j = 0; j < b.len; j++)
                c.s[i+j] += s[i] * b.s[j];
        for(int i = 0; i < c.len; i++)
        {
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator *= (const bign &b)
    {
        *this = *this * b;
        return *this;
    }
    bign operator - (const bign &b)
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator -= (const bign &b)
    {
        *this = *this - b;
        return *this;
    }
    bign operator / (const int b)
    {
        bign c;
        int f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10+s[i];
            c.s[i]=f/b;
            f=f%b;
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator / (const bign &b)
    {
        bign c, f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10;
            f.s[0] = s[i];
            while(f >= b)
            {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator /= (const bign &b)
    {
        *this  = *this / b;
        return *this;
    }
    bign operator % (const bign &b)
    {
        bign r = *this / b;
        r = *this - r*b;
        return r;
    }
    bign operator %= (const bign &b)
    {
        *this = *this % b;
        return *this;
    }
    bign operator ^ (const int & n) //大数的n次方运算
    {
        bign t(*this),ret(1);
        if(n<0) exit(-1);
        if(n==0) return 1;
        int m=n;
        while (m)
        {
            if (m&1) ret=ret*t;
            t=t*t;
            m>>=1;
        }
        return ret;
    }
    bool operator < (const bign &b)
    {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator > (const bign &b)
    {
        if(len != b.len) return len > b.len;
        for(int i = len-1; i >= 0; i--)
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        return false;
    }
    bool operator == (const bign &b)
    {
        return !(*this > b) && !(*this < b);
    }
    bool operator != (const bign &b)
    {
        return !(*this == b);
    }
    bool operator <= (const bign &b)
    {
        return *this < b || *this == b;
    }
    bool operator >= (const bign &b)
    {
        return *this > b || *this == b;
    }
    string str() const
    {
        string res = "";
        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
        return res;
    }
};

istream& operator >> (istream &in, bign &x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}

ostream& operator << (ostream &out, const bign &x)
{
    out << x.str();
    return out;
}
bign pow2[N],F[N];
int m,k;
void pre()
{
    pow2[0]=1;
    for(int i=1; i<=100; i++)pow2[i]=pow2[i-1]*2;
    F[1]=1;
    for(int i=2; i<=100; i++)F[i]=F[i-1]*2+pow2[i-1];
    return ;
}
bign cal(bign x)
{
    bign p=0;
    int num[110];
    int top=0;
    while(x.len>1||x.s[0])
    {
        num[top++]=(x.s[0]%2)?1:0;
        x=x/2;
    }
    bign ans=0;
    int j=0;
    for(int i=top-1; i>=0; i--)
    {
        if(num[i])
        {
            ans+=(F[i]+pow2[i]*j+1);
            j++;
        }
    }
    return ans;
}
bign get(bign t)
{
    bign ans=0;
    int j=0;
    for(int q=k; q>=0; q--)
    {
        bign num;
        if(q==0)
           num=j+1;
        else
           num=(F[q]+pow2[q]*j+1);
        if(t>=num)
        {
            ans=ans+pow2[q];
            t-=num;
            j=j++;
        }
    }
    return ans;
}
int main()
{
    pre();
    cin>>k>>m;
    bign l=1,r=F[k],sta=pow2[k]-l;
        while(l<r)
        {
            bign now=0;
            bign mid=(l+r)/2;
            int i;
            for(i=1; i<=m; i++)
            {
                now=get(now+mid);
                if(now>=sta)break;
                now=cal(now);
            }
            if(i>m)l=mid+1;
            else r=mid;
        }
        cout<<l<<endl;
    return 0;
}