ZOJ 3987 && 2017CCPC秦皇岛 G：Numbers（高精度+贪心）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3987

题意：

给你两个数n和m，你要将n拆成m个数，使得这m个数或和尽可能小

思路：

只要有一个数某一位是1，那么最后答案那一位就是1

所以一定要所有数中最高位的1尽可能在低位

也就是找到一个k(k=2^x)满足k*m<=n && (k+1)*m>n

这个时候ans += k，然后尽可能让m个数的最高位都为k

这样问题的规模就可以变小，递归直到0为止

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
using namespace std;
#define LL long long
int ten[4] = {1,10,100,1000};
typedef struct BigNumber
{
	int d[1200];
	BigNumber(string s)
	{
		int i, j, k, len;
		len = s.size();
		d[0] = (len-1)/4+1;
		for(i=1;i<=1199;i++)
			d[i] = 0;
		for(i=len-1;i>=0;i--)
		{
			j = (len-i-1)/4+1;
			k = (len-i-1)%4;
			d[j] += ten[k]*(s[i]-'0');
		}
		while(d[0]>1 && d[d[0]]==0)
			--d[0];
	}
	BigNumber()
	{
		*this = BigNumber(string("0"));
	}
	string toString()
	{
		int i, j, temp;
		string s("");
		for(i=3;i>=1;i--)
		{
			if(d[d[0]]>=ten[i])
				break;
		}
		temp = d[d[0]];
		for(j=i;j>=0;j--)
		{
			s = s+(char)(temp/ten[j]+'0');
			temp %= ten[j];
		}
		for(i=d[0]-1;i>0;i--)
		{
			temp = d[i];
			for(j=3;j>=0;j--)
			{
				s = s+(char)(temp/ten[j]+'0');
				temp %= ten[j];
			}
		}
		return s;
	}
}BigNumber;
BigNumber zero("0"), d, temp, mid[15];
bool operator < (const BigNumber &a, const BigNumber &b)
{
	int i;
	if(a.d[0]!=b.d[0])
		return a.d[0]<b.d[0];
	for(i=a.d[0];i>0;i--)
	{
		if(a.d[i]!=b.d[i])
			return a.d[i]<b.d[i];
	}
	return 0;
}
BigNumber operator + (const BigNumber &a, const BigNumber &b)
{
	int i, x;
	BigNumber c;
	c.d[0] = max(a.d[0], b.d[0]);
	x = 0;
	for(i=1;i<=c.d[0];i++)
	{
		x = a.d[i]+b.d[i]+x;
		c.d[i] = x%10000;
		x /= 10000;
	}
	while(x!=0)
	{
		c.d[++c.d[0]] = x%10000;
		x /= 10000;
	}
	return c;
}
BigNumber operator - (const BigNumber &a, const BigNumber &b)
{
	int i, x;
	BigNumber c;
	c.d[0] = a.d[0];
	x = 0;
	for(i=1;i<=c.d[0];i++)
	{
		x = 10000+a.d[i]-b.d[i]+x;
		c.d[i] = x%10000;
		x = x/10000-1;
	}
	while((c.d[0]>1) && (c.d[c.d[0]]==0))
		--c.d[0];
	return c;
}
BigNumber operator * (const BigNumber &a, const BigNumber &b)
{
	int i, j, x;
	BigNumber c;
	c.d[0] = a.d[0]+b.d[0];
	for(i=1;i<=a.d[0];i++)
	{
		x = 0;
		for(j=1;j<=b.d[0];j++)
		{
			x = a.d[i]*b.d[j]+x+c.d[i+j-1];
			c.d[i+j-1] = x%10000;
			x /= 10000;
		}
		c.d[i+b.d[0]] = x;
	}
	while((c.d[0]>1) && (c.d[c.d[0]]==0))
		--c.d[0];
	return c;
}
bool smaller(const BigNumber &a, const BigNumber &b, int delta)
{
	int i;
	if(a.d[0]+delta!=b.d[0])
		return a.d[0]+delta<b.d[0];
	for(i=a.d[0];i>0;i--)
	{
		if(a.d[i]!=b.d[i+delta])
			return a.d[i]<b.d[i+delta];
	}
	return 1;
}
void Minus(BigNumber &a, const BigNumber &b, int delta)
{
	int i, x;
	x = 0;
	for(i=1;i<=a.d[0]-delta;i++)
	{
		x = 10000+a.d[i+delta]-b.d[i]+x;
		a.d[i+delta] = x%10000;
		x = x/10000-1;
	}
	while((a.d[0]>1) && (a.d[a.d[0]]==0))
		--a.d[0];
}
BigNumber operator * (const BigNumber &a, int k)
{
	BigNumber c;
	c.d[0] = a.d[0];
	int i, x;
	x = 0;
	for(i=1;i<=a.d[0];i++)
	{
		x = a.d[i]*k+x;
		c.d[i] = x%10000;
		x /= 10000;
	}
	while(x>0)
	{
		c.d[++c.d[0]] = x%10000;
		x /= 10000;
	}
	while((c.d[0]>1) && (c.d[c.d[0]]==0))
		--c.d[0];
	return c;
}
BigNumber operator / (const BigNumber &a, const BigNumber &b)
{
	int i, j, temp;
	BigNumber c;
	d = a;
	mid[0] = b;
	for(i=1;i<=13;i++)
		mid[i] = mid[i-1]*2;
	for(i=a.d[0]-b.d[0];i>=0;i--)
	{
		temp = 8192;
		for(j=13;j>=0;j--)
		{
			if(smaller(mid[j], d, i))
			{
				Minus(d, mid[j], i);
				c.d[i+1] += temp;
			}
			temp /= 2;
		}
	}
	c.d[0] = max(1, a.d[0]-b.d[0]+1);
	while((c.d[0]>1) && (c.d[c.d[0]]==0))
		--c.d[0];
	return c;
}

int main(void)
{
	int T;
	string str1, str2;
	scanf("%d", &T);
	while(T--)
	{
		cin>>str1>>str2;
		BigNumber n(str1), m(str2), er("1"), ans("0"), sp("2");
		while(er*m<n+m)
			er = er*2;
		while(zero<n)
		{
			while((er*m<(n+m))==0)
				er = er/sp;
			ans = ans+er;
			if(n<er*m)
			{
				n = n/er;
				n = d;
			}
			else
				n = n-er*m;
		}
		cout<<ans.toString()<<endl;
	}
	return 0;
}