Heavy Transportation
| Time Limit: 3000MS |
|
Memory Limit: 30000K |
| Total Submissions: 5123 |
|
Accepted: 1393 |
Description
Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.
Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.
Input
The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.
Output
The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.
Sample Input
1
3 3
1 2 3
1 3 4
2 3 5
Sample Output
Scenario #1:
4
给定n个点，及m条边的最大负载，求顶点1到顶点n的最大流。
用Dijkstra算法解之，只是需要把“最短路”的定义稍微改变一下，
A到B的路长定义为路径上边权最小的那条边的长度，
而最短路其实是A到B所有路长的最大值。
//Heavy Transportation
//Dijkstra
#include <iostream>
#include<stdio.h>
using namespace std;
const int MAXS=1005;
int n;
int mat[MAXS][MAXS];
int asd[MAXS];
int s[MAXS];
int min(int a,int b)
{return a<b?a:b;}
int Dijkstra()
{
int i,j;
for(i=1;i<n;i++)
{
asd[i]=mat[0][i];
s[i]=0;
}
s[0]=1;
asd[0]=0;
for(i=0;i<n-1;i++)
{
int max=0;
int u=0;
for(j=1;j<n;j++)
{
if(s[j]==0 && asd[j]>max)
{
u=j;
max=asd[j];
}
}
if(u==0)
break;
s[u]=1;
asd[u]=max;
for(j=1;j<n;j++)
{
if (s[j]==0 && asd[j]<min(asd[u],mat[u][j]))
{
asd[j]=min(asd[u],mat[u][j]);
}
}
}
return asd[n-1];
}
int main()
{
int t,m;
int i,j;
scanf("%d",&t);
int v1,v2;
int value;
for (int s=1;s<=t;s++)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
for (j=0;j<n;j++)
{
mat[i][j]=0;
}
while (m--)
{
scanf("%d%d%d",&v1,&v2,&value);
mat[v1-1][v2-1]=mat[v2-1][v1-1]=value;
}
printf("Scenario #%d:\n%d\n\n",s,Dijkstra());
}
return 0;
}
Blocks
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 720 |
|
Accepted: 201 |
Description
Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.
Input
The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.
Output
For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.
Sample Input
2
1
2
Sample Output
2
6
Source
给定一块有n个点的木块，用四种颜色涂色，其中两种颜色只能用偶数次，求有多少种涂色方法。
一看就知是生成函数，可惜从没用过。小试身手，没想到竟然弄出来了。结果应该是对的，就是不知过程是不是可以这样写。
设四种颜色分别为w，x，y，z，其中y，z只能用偶数次，我的推导过程如下：
最后得到的公式是(2^( n - 1 ))(2^(n-1)+1)
注意到10007是素数,由费尔马定理,可以先把n-1mod（10007-1）,减小计算量,剩下的就是快速取幂了.
#include <iostream>
using namespace std;
const int mod=10007;
int pow(int n)
{
if(n==0)
return 1;
if(n&1)
{
return (pow(n-1)<<1)%mod;
}
else
{
int temp=pow(n>>1);
return (temp*temp)%mod;
}
}
int main(int argc, char *argv[])
{
int t,n,temp;
cin>>t;
while(t--)
{
cin>>n;
temp=pow((n-1)%(mod-1));
cout<<(temp*(temp+1))%mod<<endl;
}
return 0;
}
//由于近日POJ登不上，上面的代码未曾提交过
Euclid's Game
|
Time Limit: 1000MS |
|
Memory Limit: 65536K |
|
Total Submissions: 4525 |
|
Accepted: 1849 |
Description
Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7):
25 7
11 7
4 7
4 3
1 3
1 0
an Stan wins.
Input
The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.
Output
For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed.
Sample Input
34 12
15 24
0 0
Sample Output
Stan wins
Ollie wins
Source
Waterloo local 2002.09.28
给定两堆石子,二人轮流取子,要求只能从石子数目较大的那一堆取子,取子的数目只能是另一堆石子数目的倍数.最终使得某一堆数目为零的一方为胜.
首先,容易看出,对于每一个局面,要么是先手必胜,要么是后手必胜,最终结果完全由当前局面完全确定.
另外,可以简单罗列一下先手必胜和必败的几种局面(两堆石子初始数目都大于零):
1,有一堆石子数目为一,先手必胜, 1,4, 1,2.
2,两堆石子数目差一,且两堆石子数目都不为一,先手必败(只能使后手面对必胜的局面),如 3,4 5,6 .
3,如果数目较大的那一堆是数目较小那一堆的2倍加减一,且不是上面两种局面,先手必胜,2,5 3,5 3,7.
可是上面这些信息对于解决这个问题还是有一些困难.
再进一步试算数目较小的石子,可以发现,当两堆数目相差较大时,总是先手必胜.
事实上,进一步探讨可以发现下面的结论:
1,N<2*M-1时,先手别无选择,只能使之变为 N-M,M 局面,(易见)如3,5 5,7 7,4...
2,设两堆石子数目为N,M(N>M>0,且N,M互质),则若N>=2*M-1,且N - M ! =1时,先手必胜.要求M,N互质是因为对于M,N有公因数的情形,可以同时除以其公因数而不影响结果.
简单说明一下上面结论2的由来. N>=2*M-1时,先手可使之变为 N%M,M 或N%M+M,M两种局面之一,其中有且只有一个必败局面。注意到如果N%M,M不是必败局面，那么N%M+M,M就是必败局面，因为面对N%M+M,M这个局面，你别无选择，只能在前一堆中取M个使对方面对必胜局面(结论1 )。
据此可设计算法如下:
1.M,N先同时除以它们的最大公因数.(M<N)
2,如果M==0,则返回零;
3,如果M==1,则返回一;
4,如果N>=M*2-1,则返回一
5,令N=M,M=N-M,递归处理
#include <iostream>
using namespace std;
long long gcd(long long a,long long b)
{
if(a==0)
return b;
return gcd(b%a,a);
}
long long Eu(long long m,long long n)
{
if(m==1)
return 1;
if(n-m==1 && m)
return 0;
if(n>=m*2-1)
return 1;
return !Eu(n%m,m);
}
int main()
{
long long m,n,temp;
while (cin>>m>>n && (m||n))
{
long long g=gcd(m,n);
m/=g;
n/=g;
if(m>n)
{
temp=m;
m=n;
n=temp;
}
if(Eu(m,n))
cout<<"Stan wins"<<endl;
else
cout<<"Ollie wins"<<endl;
}
return 0;
}
Rank List
|
Time Limit: 10000MS |
|
Memory Limit: 65536K |
|
Total Submissions: 6561 |
|
Accepted: 2091 |
Description
Li Ming is a good student. He always asks the teacher about his rank in his class after every exam, which makes the teacher very tired. So the teacher gives him the scores of all the student in his class and asked him to get his rank by himself. However, he has so many classmates, and he can’t know his rank easily. So he tends to you for help, can you help him?
Input
The first line of the input contains an integer N (1 <= N <= 10000), which represents the number of student in Li Ming’s class. Then come N lines. Each line contains a name, which has no more than 30 letters. These names represent all the students in Li Ming’s class and you can assume that the names are different from each other.
In (N+2)-th line, you'll get an integer M (1 <= M <= 50), which represents the number of exams. The following M parts each represent an exam. Each exam has N lines. In each line, there is a positive integer S, which is no more then 100, and a name P, which must occur in the name list described above. It means that in this exam student P gains S scores. It’s confirmed that all the names in the name list will appear in an exam.
Output
The output contains M lines. In the i-th line, you should give the rank of Li Ming after the i-th exam. The rank is decided by the total scores. If Li Ming has the same score with others, he will always in front of others in the rank list.
Sample Input
3
Li Ming
A
B
2
49 Li Ming
49 A
48 B
80 A
85 B
83 Li Ming
Sample Output
1
2
Source
POJ Monthly,Li Haoyuan
给定每个人的成绩，查询某一人的名次。
用MAP建立人名和成绩的对应关系，用cnt数组（最多5000个元素）记录成绩为某个分数的人数，不过由于总人数较少(最多只有10000人),直接遍历也不比建立计数排序数组多用多少时间,计数排序的优势并不显著.
用hash函数或者二分查找也应该能解决这个问题.
/**//*Source Code
Problem: 2153 User: y09
Memory: 1236K Time: 1204MS
Language: C++ Result: Accepted
Source Code */
#include <iostream>
#include<string>
#include<map>
using namespace std;
int main(int argc, char *argv[])
{
int n,m;
int i,j;
char str[200];
string str1;
map<string ,int>score;
scanf("%d",&n);
getchar();
for (i=0;i<n;i++ )
{
gets(str);
str1=str;
score[str1]=0;
}
int cnt[5005]={0};
scanf("%d",&m);
string li="Li Ming";
int rank=0;
int s=0;
int temp=0;
int temp2=0;
int num;
for(int k=0;k<m;k++)
{
for(i=0;i<n;i++)
{
scanf("%d",&num);
getchar();
gets(str);
str1=str;
temp2=score[str1];
score[str1]=num+temp2;
cnt[num+temp2]++;
}
s=score[li];
rank=1;
temp+=100;
for(i=temp;i>s;i--)
{
rank+=cnt[i];
cnt[i]=0;
}
for(i=s;i>=0;i--)
cnt[i]=0;
printf("%d\n",rank);
}
return 0;
}
Binary Stirling Numbers
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 1040 |
|
Accepted: 346 |
Description
The Stirling number of the second kind S(n, m) stands for the number of ways to partition a set of n things into m nonempty subsets. For example, there are seven ways to split a four-element set into two parts:
{1, 2, 3} U {4}, {1, 2, 4} U {3}, {1, 3, 4} U {2}, {2, 3, 4} U {1}
{1, 2} U {3, 4}, {1, 3} U {2, 4}, {1, 4} U {2, 3}.
There is a recurrence which allows to compute S(n, m) for all m and n.
S(0, 0) = 1; S(n, 0) = 0 for n > 0; S(0, m) = 0 for m > 0;
S(n, m) = m S(n - 1, m) + S(n - 1, m - 1), for n, m > 0.
Your task is much "easier". Given integers n and m satisfying 1 <= m <= n, compute the parity of S(n, m), i.e. S(n, m) mod 2.
Example
S(4, 2) mod 2 = 1.
Task
Write a program which for each data set:
reads two positive integers n and m,
computes S(n, m) mod 2,
writes the result.
Input
The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 200. The data sets follow.
Line i + 1 contains the i-th data set - exactly two integers ni and mi separated by a single space, 1 <= mi <= ni <= 10^9.
Output
The output should consist of exactly d lines, one line for each data set. Line i, 1 <= i <= d, should contain 0 or 1, the value of S(ni, mi) mod 2.
Sample Input
1
4 2
Sample Output
1
Source
判断第二类斯特灵数模 2 的余数。
在刘汝佳的黑书上有详细解答，基本思路是枚举数值较小的斯特灵数，从中寻找规律。
下面这幅图是从维基百科截出来的，有一个二进制斯特灵数与组合数的转化公式。而组合数模二的余数就很容易了。
我们知道，组合数C（N,M）=N ! / M ! /(N-M)!,因而只需求得阶乘质因数分解式中二的重数即可解决问题。
而N ！质因数分解后2的重数可用下式来计算之。
K=N/2+N/2^2+N/2^3+....
上式的除法全是下取整。（可参见任何一本初等数论课本，如北大潘承洞编的那本《初等数论》）。
这样，这个问题就迎刃而解。
另外，有一点说明的是上面那个图形，就是分形几何中一个很重要的例子——谢彬斯基垫片。杨辉三角也有类似的形状。
这是我用MATLAB作的一个杨辉三角的二进制图形。
#include <stdio.h>
int main(int argc, char *argv[])
{
int n,m;
int z,w1,w2;
int t;
int a,b,c;
scanf("%d",&t);
while (t--)
{
scanf("%d%d",&n,&m);
z=n-(m+2)/2;
w1=(m-1)/2;
w2=z-w1;
a=0;
while (z)
{
z>>=1;
a+=z;
}
b=0;
while (w1)
{
w1>>=1;
b+=w1;
}
c=0;
while (w2)
{
w2>>=1;
c+=w2;
}
printf("%d\n",(a-b-c)==0);
}
return 0;
}
Matrix Multiplication
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 11924 |
|
Accepted: 2408 |
Description
You are given three n × n matrices A, B and C. Does the equation A × B = C hold true?
Input
The first line of input contains a positive integer n (n ≤ 500) followed by the the three matrices A, B and C respectively. Each matrix's description is a block of n × n integers.
It guarantees that the elements of A and B are less than 100 in absolute value and elements of C are less than 10,000,000 in absolute value.
Output
Output "YES" if the equation holds true, otherwise "NO".
Sample Input
2
1 0
2 3
5 1
0 8
5 1
10 26
Sample Output
YES
Hint
Multiple inputs will be tested. So O(n3) algorithm will get TLE.
Source
给定矩阵A和B，判断矩阵C是不是它们的乘积。
题目明确表示直接判断会超时，而Strass和直接相乘的O（n^3）效果相差不多。
因而采用随机化方法，按我自己的想法，随机测试C中的若干元素，以确定结果，看了讨论区，才发现有更加“专业”的办法。
随机生成行向量I，则若A*B=C，那么必有I*A*B=I*C；反之，不一定成立，算法的随机性正体现在这里。
用一个必要不充分条件来判断结果的正确性，比盲目测试效果往往要好得多。
这个必要条件判断结果的时间复杂度是O（N^2）的，这是题目输入数据量可以接受的。
/**//*Source Code
Problem: 3318 User: y09
Memory: 3080K Time: 1063MS
Language: C Result: Accepted
Source Code */
#include <stdio.h>
#include <time.h>
#include<stdlib.h>
int main()
{
int n;
int i,j;
int mat1[500][500];
int mat2[500][500];
int mat3[500][500];
__int64 te1[500]={0};
__int64 te2[500]={0};
__int64 te3[500]={0};
__int64 te4[500]={0};
time_t t;
srand((unsigned) time(&t));
scanf("%d",&n);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&mat1[i][j]);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&mat2[i][j]);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&mat3[i][j]);
for(i=0;i<n;i++)
te1[i]=rand()%100;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
te2[i]+=te1[j]*mat1[j][i];
for(i=0;i<n;i++)
for(j=0;j<n;j++)
te3[i]+=te2[j]*mat2[j][i];
for(i=0;i<n;i++)
for(j=0;j<n;j++)
te4[i]+=te1[j]*mat3[j][i];
for(i=0;i<n;i++)
if(te3[i]!=te4[i])
{
puts("NO");
return 0;
}
puts("YES");
return 0;
}
Mobile phones
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 7087 |
|
Accepted: 3030 |
Description
Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.
Input
The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.
The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30
Output
Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.
Sample Input
0 4
1 1 2 3
2 0 0 2 2
1 1 1 2
1 1 2 -1
2 1 1 2 3
3
Sample Output
3
4
Source
一维树状数组用一维数组来存储部分元素的和，二维树状数组只需用二维数组来存储即可，获得和，修正的函数同一维数组差别不大。
/**//*Source Code
Problem: 1195 User: y09
Memory: 4956K Time: 579MS
Language: C++ Result: Accepted
Source Code */
#include <stdio.h>
const int MAX=1200;
int c[MAX][MAX];
int n;
int LowBit(int t)
{
return t&(t^(t-1));
}
int Sum(int endx,int endy)
{
int sum=0;
int temp=endy;
while(endx>0)
{
endy=temp;//注意记录endy的值，本人在此出错，找半天错误不得
while (endy>0)
{
sum+=c[endx][endy];
endy-=LowBit(endy);
}
endx-=LowBit(endx);
}
return sum;
}
void plus(int posx,int posy,int num)
{
int temp=posy;
while (posx <=n)
{
posy=temp;
while(posy<=n)
{
c[posx][posy]+=num;
posy+=LowBit(posy);
}
posx+=LowBit(posx);
}
}
int GetSum(int l,int b,int r,int t)
{
return Sum(r,t)-Sum(r,b-1)-Sum(l-1,t)+Sum(l-1,b-1);
}
int main()
{
int I;
int x,y,a;
int l,b,r,t;
while(scanf("%d",&I))
{
switch (I)
{
case 0:
scanf("%d",&n);
break;
case 1:
scanf("%d%d%d",&x,&y,&a);
plus(x+1,y+1,a);
break;
case 2:
scanf("%d%d%d%d",&l,&b,&r,&t);
printf("%d\n",GetSum(l+1,b+1,r+1,t+1));
break;
case 3:
return 0;
}
}
return 0;
}
Cipher
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 12776 Accepted: 3194
Description
Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is based on secret keys. They chose the secret key at their last meeting in Philadelphia on February 16th, 1996. They chose as a secret key a sequence of n distinct integers, a1 ; . . .; an, greater than zero and less or equal to n. The encoding is based on the following principle. The message is written down below the key, so that characters in the message and numbers in the key are correspondingly aligned. Character in the message at the position i is written in the encoded message at the position ai, where ai is the corresponding number in the key. And then the encoded message is encoded in the same way. This process is repeated k times. After kth encoding they exchange their message.
The length of the message is always less or equal than n. If the message is shorter than n, then spaces are added to the end of the message to get the message with the length n.
Help Alice and Bob and write program which reads the key and then a sequence of pairs consisting of k and message to be encoded k times and produces a list of encoded messages.
Input
The input file consists of several blocks. Each block has a number 0 < n <= 200 in the first line. The next line contains a sequence of n numbers pairwise distinct and each greater than zero and less or equal than n. Next lines contain integer number k and one message of ascii characters separated by one space. The lines are ended with eol, this eol does not belong to the message. The block ends with the separate line with the number 0. After the last block there is in separate line the number 0.
Output
Output is divided into blocks corresponding to the input blocks. Each block contains the encoded input messages in the same order as in input file. Each encoded message in the output file has the lenght n. After each block there is one empty line.
Sample Input
10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0
Sample Output
BolHeol b
C RCE
Source
Central Europe 1995
给定1~n的置换F,求其变换m次的变换F^m.
先找到循环节,再用m对循环节的长度取模即可.
#include <iostream>
using namespace std;
int main()
{
const int MAX=300;//最大长度
char str[MAX];//读入串
int n;//变换的长度
int data[MAX]={0};//存放原始变换
int used[MAX]={0};//标志数组
int cir[MAX][MAX]={0};//每个循环节的成员
int num[MAX]={0};//循环节对应长度
int cnt=0;//循环节的个数
int time=0;//变换次数
int change[MAX]={0};//原始循环变换time次之后的变换
char res[MAX]={0};//变换之后的字符串
int i,j;
while(cin>>n && n)
{
memset(used,0,sizeof(used));
memset(num,0,sizeof(num));
for(i=1;i<=n;i++)
cin>>data[i];
cnt=0;//计数循环节个数
for(i=1;i<=n;i++)
{
if(used[i]==0)
{
used[i]=1;
int temp=data[i];
cir[cnt][num[cnt]]=temp;
num[cnt]=1;
while(used[temp]==0)//获得循环节
{
used[temp]=1;
temp=data[temp];
cir[cnt][num[cnt]++]=temp;
}
cnt++;
}
}
while(cin>>time && time)//读入变换次数
{
memset(res,0,sizeof(res));
memset(str,0,sizeof(str));
gets(str);
int len=strlen(str);
for(i=len;i<=n;i++)//位数不足n,补空格
str[i]=' ';
//获得变换
for(i=0;i<cnt;i++)
{
for(j=0;j<num[i];j++)
{
change[cir[i][j]]=cir[i][(j+time)%num[i]];
}
}
//对读入数据变换,获得结果
for(i=1;i<=n;i++)
{
res[change[i]]=str[i];
}
cout<<res+1<<endl;
}
cout<<endl;
}
return 0;
}
题意简单,就是找四个数的和为零.先把前两列的和算出来,O(n^2),存到hash表中,再把后两列的两两和算出来,在hash表中找相反数.用线性探测法就可以解决该问题了.
/**//*Source CodeProblem: 2785 User: y09shendazhi
Memory: 160132K Time: 2610MS
Language: G++ Result: Accepted
Source Code*/
#include <iostream>
#include<stdio.h>
using namespace std;
const int size=20345677;
int hash[size];
int sum[size];
const int key=1777;
const int MAX=1000000000;
void Insert(int num)
{
int temp=num;
num=(num+MAX)%size;
while(hash[num]!=MAX && hash[num]!=temp)
num=(key+num)%size;
hash[num]=temp;
sum[num]++;
}
int Find(int num)
{
int temp=num;
num=(num+MAX)%size;
while(hash[num]!=MAX && hash[num]!=temp)
num=(num+key)%size;
if(hash[num]==MAX)
return 0;
else
return sum[num];
}
int main()
{
int n;
int i,j;
int a[4005],b[4005],c[4005],d[4005];
scanf("%d",&n);
int ans=0;
for(i=0;i<n;i++)
{
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
}
for(i=0;i<size;i++)
{
hash[i]=MAX;
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
Insert(a[i]+b[j]);
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
ans+=Find(-(c[i]+d[j]));
}
printf("%d\n",ans);
return 0;
}
读半天不解其意,傻算了半天才把sample凑出来.其实很简单,连高精度都没有.第一类斯特灵数S(n,m)就是把n元集合分成m部的个数,有递推关系S(n,m)=S(n-1,m-1)+mS(n-1,m).所求还要全排列一下.再乘以m!就可以了累加1~B个数全部用上,就是结果.F(B)=sum(C(B,i)*(Sum(Stir( i,j )* j ! ) ) )就是结果.其中下标i从1变到B,j从1变到i.
/**//*Source CodeProblem: 3088 User: y09shendazhi
Memory: 164K Time: 0MS
Language: C++ Result: Accepte/
Source Code*/
#include <iostream>
using namespace std;
// /n\
// | |
// \m/
__int64 comb(__int64 m,__int64 n)
{
if(n<m)
return 0;
__int64 result=1;
m=m<n-m?m:n-m;
for(__int64 i=1;i<=m;i++)
result=(result*(n-m+i))/i;
return result;
}
__int64 fun(__int64 n)
{
__int64 ans=1;
for(__int64 i=1;i<=n;i++)
ans*=i;
return ans;
}
int main(__int64 argc, char *argv[])
{
__int64 i,j,k;
__int64 stir[15][15]={0};
for(i=1;i<12;i++)
{
stir[i][1]=1;
for(j=2;j<i;j++)
{
stir[i][j]=stir[i-1][j-1]+j*stir[i-1][j];
}
stir[i][i]=1;
}
__int64 ans[12]={0,1};
for(i=2;i<12;i++)
{
for(j=1;j<=i;j++)
{
__int64 temp=0;
for(k=1;k<=j;k++)
{
temp+=stir[j][k]*fun(k);
}
ans[i]+=temp*comb(j,i);
}
}
__int64 in=0;
__int64 t=0;
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&in);
printf("%I64d %I64d %I64d\n",i+1,in,ans[in]);
}
return 0;
}