千题目标Category

LightOJ – 1045 K进制的位数

Factorial of an integer is defined by the following function

f(0) = 1

f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input
Input starts with an integer T (≤ 50000), denoting the
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线性同余方程组的合并

类似于

[latex][/latex]

a_ix\equiv b_i\mod m_i

的方程组就是线性同余方程组
根据同余的可加性，可令

x\equiv B \mod lcm(m_i)

所以答案就转化为了求B和lcm

a_1x \equiv b_1 \mod m1

x\equiv (b_1/\gcd(a_1,m_1))*(a_1^{-1}/gcd(a_1,m_1)) \mod (m_1/gcd(a_1,m_1))

设

x \equiv B_1 \mod m1

x=B_1+m_1*t 此时的m1*t是未知的

带入第二个方程,并整理

a(B_1+m_1t) \equiv b_2 \mod m_2

am_1t \equiv b_2-aB_1 \mod m_2

解出t

t \equiv (b_2-aB_1)/gcd(am_1,m_2) * (am_1)^{-1}/gcd(am_1,m_2) \mod (m_2)/gcd(am_1,m_2)

把t带入

x=B_1+m_1*t

[latex]此时的x在\mod m1和\mod m2的意义下都是成立的。B1_+m_1*t变成了新的B_2[/latex]

x\equiv B_2 \mod lcm(m_1,m_2)

x=B_2+lcm*t

再带入第三个方程即可向下合并

//返回一个b，m对 
pair<int, int> inner_congruence(const vector<int> &A,const vector<int> &B,const vector<int> &M)
{
	int x=0,m=1;
	for(int i-9;i<A.size();i++)
	{
		int a=A[i]*m,b=B[i]-A[i]*x,d=gcd(a,M[i]);
		if(b%d!=0) return make_pair(0,-1);
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【容斥原理】Arrange the Numbers LightOJ – 1095

Consider this sequence {1, 2, 3 ... N}, as an initial sequence of first N natural numbers. You can rearrange this sequence in many ways. There will be a total of N! arrangements. You have to calculate the number of arrangement of first N natural numbers, where in first M positions; exactly K numbers are in their initial position.

For Example, N = 5, M = 3, K = 2

You should count this arrangement {1, 4, 3, 2, 5},
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【矩阵快速幂】nth Term LightOJ – 1096

You have to find the nth term of the following function:

f(n) = a * f(n-1) + b * f(n-3) + c, if(n > 2)

     = 0, if(n ≤ 2)

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains four integers n (0 ≤ n ≤ 108), a b c (1 ≤ a, b, c ≤ 10000).

Output
For each case, print the case number and f(n) modulo 10007.

Sample Input
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【坑】高数满分—pow的问题

Description

众所周知，qm的高数总是满分。这一天数学白痴锦鱼拿着一道数学题来问qm，qm不屑做这种简单的题，于是把这道题丢给你来完成。题目是这样的：给定a、b、c三个整数，求最大的整数x，满足xa+b*x<=c,其中xa表示x的a次方，b*x表示b乘以x。

Input

第一行输入三个整数a(1<=a<=10),b(0<=b<=1018),c(0<=c<=1018)，分别对应题目中a、b、c三个整数。

Output

输出一个整数，表示所求的最大的x。

Sample Input

2 3 4

Sample Output

1
这题虽然是一道二分，但由于x^a可能过于大，所以x每次要去pow(c,1.0/a)避免爆炸。
xyjj说pow在计算longlong的时候会出现各种奇怪的问题（导致我WA了无数次），后来网上查了一下说是因为pow是计算的double所以会有精度丢失的问题，所以二分里面的pow要自己写……………….
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a,b,c;
ll qkpow(ll x,int p)
{
	if(p==0) return 1;
	if(p==1) return x; 
	
	ll y=qkpow(x,p>>1);
	
	y=y*y;
	
	if(p&1) 
	y=y*x;
	
	return y;
}
int main()
{
	cin>>a>>b>>c;
	ll l=0,r=(ll)pow((double)c,1.0/a);
	//if(b!=0) r=c/b;
	//r=min(r,(ll)pow(c,1.0/a));
	while(r-l>1)
	{
		ll mid=(r+l)>>1;
		if(qkpow(mid,a)+b*mid<=c)
		{
			l=mid;	
		}
		else
		{
			r=mid;
		}
	}	
	if(qkpow(r,a)+r*b<=c) cout<<r;
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