Project Euler 48-Self powers

Self powers

Problem 48

The series, 11 + 22 + 33 + ... + 1010 = 10405071317.

Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.

 

Answer:
9110846700
Completed on Fri, 5 Apr 2013, 12:33

Go to the thread for problem 48 in the forum.

 

ans=0
i=1
F=10**10
for i in range(1,1001):
p=1
for j in range(i): p=p*i%F
ans=(ans+p)%F
print i
print ans

Project Euler 3-Largest prime factor

Largest prime factor

Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

 

Answer:
6857
Completed on Fri, 5 Apr 2013, 12:14

Go to the thread for problem 3 in the forum.

 

import math
n=600851475143
i=2
print math.sqrt(n)
while i< =math.sqrt(n): if (n%i==0): while (n%i==0): n=n/i print i; i=i+1 print n

Project Euler 2-Even Fibonacci numbers

Even Fibonacci numbers

Problem 2

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Answer:
4613732
Completed on Fri, 5 Apr 2013, 06:37

Go to the thread for problem 2 in the forum.

 Even(是偶数的意思……(─.─|||

f=[1,1,1]
i=2
ans=0
while f[i]+f[i-1]<4000000:
i=i+1
f.append(f[i-1]+f[i-2])
if (f[i]%2==0): ans=ans+f[i]
print sum(f)-1,f,ans

UVA 11538(Chess Queen-矩阵对角线长度)

Problem A
Chess Queen
Input:
Standard Input

Output: Standard Output

 

王后互相攻击的前提是在一行,一列,或一对角线。:

 

在 (NxM) 的棋盘上 2 王后互相攻击,求方案数.

 

Input

 

输入数据不超过 5000 行 ,每行为M and N (0< M, N£106) ,数据以0 0结尾.

 

Output

 

每行一个整数表示方案数,保证它在u64范围内.

 

Sample Input                              Output for Sample Input

2 2

100 223

2300 1000

0 0

12

10907100

11514134000                                                                        

 

Problemsetter: Shahriar Manzoor

Special Thanks to: Mohammad Mahmudur Rahman

首先,一个矩形的长宽若为m,n(m>=n)

那么它一个方向的对角线应为1..(n-1)各2条,n有(m-n+1)条

知道这个的化,本题就转化为,在一列一行或一对角线任取2点,有几种取法。

#include<cstdio>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cstdlib>
#include<cstring>
using namespace std;
#define MAXN (1000000+10)
unsigned long long n,m;
int main()
{
	while (cin>>n>>m&&n&&m)
	{
		if (n>m) swap(n,m);
		cout<<n*m*(n+m-2)+2*n*(n-1)*(3*m-n-1)/3<<endl;
	}
	return 0;
}



BZOJ 3098(Hash Killer II-生日攻击)

3098: Hash Killer II

Time Limit: 5 Sec  Memory Limit: 128 MBSec  Special Judge
Submit: 66  Solved: 22
[Submit][Status][Discuss]

Description

这天天气不错,hzhwcmhf神犇给VFleaKing出了一道题:
给你一个长度为N的字符串S,求有多少个不同的长度为L的子串。
子串的定义是S[l]、S[l + 1]、... S[r]这样连续的一段。
两个字符串被认为是不同的当且仅当某个位置上的字符不同。

VFleaKing一看觉得这不是Hash的裸题么!于是果断写了哈希 + 排序。
而hzhwcmhf神犇心里自然知道,这题就是后缀数组的height中 < L的个数 + 1,就是后缀自动机上代表的长度区间包含L的结点个数,就是后缀树深度为L的结点的数量。
但是hzhwcmhf神犇看了看VFleaKing的做法表示非常汗。于是想卡掉他。

VFleaKing使用的是字典序哈希,其代码大致如下:
u64 val = 0;
for (int i = 0; i < l; i++)
 val = (val * base + s[i] - 'a') % Mod;
u64是无符号int64,范围是[0, 2^64)。
base是一个常量,VFleaKing会根据心情决定其值。
Mod等于1000000007。
VFleaKing还求出来了base ^ l % Mod,即base的l次方除以Mod的余数,这样就能方便地求出所有长度为L的子串的哈希值。
然后VFleaKing给哈希值排序,去重,求出有多少个不同的哈希值,把这个数作为结果。
其算法的C++代码如下:

typedef unsigned long long u64;

const int MaxN = 100000;

inline int hash_handle(const char *s, const int &n, const int &l, const int &base)
{
 const int Mod = 1000000007;

 u64 hash_pow_l = 1;
 for (int i = 1; i <= l; i++)
  hash_pow_l = (hash_pow_l * base) % Mod;

 int li_n = 0;
 static int li[MaxN];

 u64 val = 0;
 for (int i = 0; i < l; i++)
  val = (val * base + s[i] - 'a') % Mod;
 li[li_n++] = val;
 for (int i = l; i < n; i++)
 {
  val = (val * base + s[i] - 'a') % Mod;
  val = (val + Mod - ((s[i - l] - 'a') * hash_pow_l) % Mod) % Mod;
  li[li_n++] = val;
 }

 sort(li, li + li_n);
 li_n = unique(li, li + li_n) - li;
 return li_n;
}

hzhwcmhf当然知道怎么卡啦!但是他想考考你。

Input

没有输入。

Output

你需要输出一组数据使得VFleaKing的代码WA掉。我们会使用Special Judge检查你的结果的正确性。
第一行两个用空格隔开的数n、l。
第二行是一个长度为n的字符串。只能包含'a'~'z'。
需要保证1 <= n <= 10^5, 1 <= l <= n,
不符合以上格式会WA。
不要有多余字符,很可能导致你WA。

Sample Input

没有

Sample Output

8 4

buaabuaa

(当然这个输出是会WA的)

HINT

如果一个房间里有23个或23个以上的人,那么至少有两个人的生日相同的概率要大于50%。

Source

生日攻击:如果你在n个数中随机选数,那么最多选√n次就能选到相同的数(不考虑Rp broken)

同样的,这题的Hash值在0到1000000007.

那就要选差不多10^5次

唯一注意的是l要取大,使得方案数超过Mod

否则就不可能有2个数有相同的Hash值

#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXN (100000)
int n=MAXN,l=13;
int main()
{
	cout<<n<<' '<<l<<endl;
	for(int i=1;i<=n;i++) cout<<char(rand()%26+'a');
	cout<<endl;
	return 0;
}




POJ 2484(博弈-对称博弈)

A Funny Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3345   Accepted: 1960

Description

Alice and Bob decide to play a funny game. At the beginning of the game they pick n(1 <= n <= 106) coins in a circle, as Figure 1 shows. A move consists in removing one or two adjacent coins, leaving all other coins untouched. At least one coin must
be removed. Players alternate moves with Alice starting. The player that removes the last coin wins. (The last player to move wins. If you can't move, you lose.) 
 

Note: For n > 3, we use c1, c2, ..., cn to denote the coins clockwise and if Alice remove c2, then c1 and c3 are NOT adjacent! (Because there is an empty place between c1 and c3.) 

Suppose that both Alice and Bob do their best in the game. 
You are to write a program to determine who will finally win the game.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (1 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input. 

Output

For each test case, if Alice win the game,output "Alice", otherwise output "Bob". 

Sample Input

1
2
3
0

Sample Output

Alice
Alice
Bob

Source

POJ Contest,Author:Mathematica@ZSU

博弈第一招——对称博弈。

思路:无论对方怎么取,我都取与它对称(或者相反/一样的)

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
	int n;
	while(scanf("%d",&n)&&n)
	{
		if (n>=3) cout<<"Bob"<<endl;
		else cout<<"Alice"<<endl;
	}
	return 0;
}

POJ 2362(Square-搜索剪枝1-相对顺序)

Language:
Square
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 17066   Accepted: 5878

Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

Sample Input

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

Sample Output

yes
no
yes

Source

《搜索是怎样剪枝的-1》

1.只要找到3条边。

2.从大到小顺序找。

3.每次搜边时要确定边的相对顺序唯一(直接从TLE→秒)


#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXM (20+10)
int tt,n,a[MAXM],cnt,len;
bool b[MAXM],flag;
bool dfs(int l,int x,int pre)
{
//	cout<<l<<' '<<x<<' '<<kth<<endl;
	if (x==len) {l++;x=0;pre=n-1;}
	if (l==4)
	{
		return 1;
	}
	for(int i=pre-1;i;i--)
		if (!b[i]&&x+a[i]<=len)
		{
			b[i]=1;
			if (dfs(l,x+a[i],i)) return 1;
			b[i]=0;
		//	if (!x) return 0;
		}
	return 0;
}
int main()
{
	scanf("%d",&tt);
	while (tt--)
	{
		cnt=0;
		scanf("%d",&n);
		for (int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			cnt+=a[i];b[i]=0;
		}
		sort(a+1,a+1+n);
		if (n<4||cnt%4||a[n]>cnt/4)
		{
			printf("non");continue;
		}
		b[n]=1;len=cnt/4;
		if (dfs(1,a[n],n))
		{
			printf("yesn");
		}
		else printf("non");
	}
	return 0;
}


POJ 2553(The Bottom of a Graph-缩点求出度)

Language:
The Bottom of a Graph
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 7308   Accepted: 3003

Description

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges.
Then G=(V,E) is called a directed graph. 
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1).
Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+1)
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from vv is also reachable from w. The bottom of a graph is the subset of
all nodes that are sinks, i.e.,bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.

Input

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer
numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v1,w1,...,ve,we with
the meaning that (vi,wi)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 2 3 3 1
2 1
1 2
0

Sample Output

1 3
2

Source

Tarjen缩点统计出度。


#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXN (5000+10)
#define MAXM (1000000+10)
int n,m,h[MAXN],t[MAXN],dfs[MAXN],tim,kind,s[MAXN],tot;
bool b[MAXN];
int edge[MAXM],next[MAXM],pre[MAXN],size;
int x[MAXM],y[MAXM],numk[MAXN];
void addedge(int u,int v)
{
	edge[++size]=v;
	next[size]=pre[u];
	pre[u]=size;
}
void tar(int u)
{
	dfs[u]=t[u]=++tim;b[u]=1;s[++tot]=u;
	for (int p=pre[u];p;p=next[p])
	{
		int &v=edge[p];
		if (!b[v]) tar(v),dfs[u]=min(dfs[u],dfs[v]);
		else if (!h[v]) dfs[u]=min(dfs[u],t[v]);  //保证指向祖先
	}
	if (dfs[u]==t[u])
	{
		kind++;
		while (s[tot]!=u) h[s[tot--]]=kind,numk[kind]++;
		h[s[tot--]]=kind;numk[kind]++;

	}
}
int outdegree[MAXN];
int main()
{
	while (scanf("%d",&n)&&n)
	{
		scanf("%d",&m);
		tot=size=tim=kind=0;
		memset(h,0,sizeof(h));
		memset(t,0,sizeof(t));
		memset(dfs,0,sizeof(dfs));
		memset(s,0,sizeof(s));
		memset(pre,0,sizeof(pre));
		memset(b,0,sizeof(b));
		memset(numk,0,sizeof(numk));
		memset(outdegree,0,sizeof(outdegree));

		for (int i=1;i<=m;i++)
		{
			scanf("%d%d",&x[i],&y[i]);
			addedge(x[i],y[i]);
		}
		for (int i=1;i<=n;i++) if (!b[i]) tar(i);
//		for (int i=1;i<=n;i++) cout<<h[i]<<' ';cout<<endl;
		for (int i=1;i<=m;i++)
		{
			if (h[x[i]]^h[y[i]]) outdegree[h[x[i]]]++;
		}
		/*
		int ans=0;
		for (int i=1;i<=kind;i++)
			if (!outdegree[i]) ans+=numk[i];
		cout<<ans<<endl;
		*/
		bool flag=0;
		for (int i=1;i<=n;i++)
			if (!outdegree[h[i]])
			{
				if (flag) printf(" ");else flag=1;printf("%d",i);
			}


	//	cout<<ans<<endl;
		cout<<endl;
	}
	return 0;
}