标签：soj1942 foto 数论 排列组合 快速排列组合
Description
The election campaign just started, so PSOS decided to make some propagation. One of the representatives came with a great idea - he proposes to make an photography of their Parliament Club. Unfortunatelly, even after many briefings, the representatives are still not able to agree upon an ordering of people in the photography. Moreover, there are a lot of representatives and it is not possible to have all of them in a single picture. The situation became critical. In the end, the representatives decided they will make one photo of every possible combination of people and their ordering. Different photographs will be used for a large number of billboards PSOS plan to use. To make things more clear, every person gets a Unique Identification Number (UIN). Every picture can be then described as the succession of several UINs x1, x2, ... xk, in whichxi is UIN of the i-th person in the picture x. Now we can sort all the possible photographs (combinations) to a single succession. The ordering of combinations of the same length (photographs with the same number of people in it) is defined as follows: the combination p is greater than the combination q if there exists any i such as that pj = qj for every j < i, and pi > qi. Your goal is to find the right place for a given picture among all possible photographs.
At the first line there is a positive integer N stating the number of assignments to follow. Each assignment consists of exactly two lines. At the first line of each assignment, there are two integers n and k, 1 <= k <= n <= 12 stating the total number of representatives (n) and the number of them which can fit into a single picture (k). At the second line of the assignment, there is exactly k positive numbers x1, x2, ... xk, each of them 1 <= xi <= n. No number can appear more than once on this line.
For each assignment, output the text "Variace cislo I ma poradove cislo J." (Combination #I is J-th in sequence). Fill the number of assignment instead of I (starting with one), and the number of the given photograph among all possible combinations after ordering, instead of J (also starting with one).
4 1 1 1 5 1 4 3 3 1 2 3 5 3 5 3 1
Variace cislo 1 ma poradove cislo 1. Variace cislo 2 ma poradove cislo 4. Variace cislo 3 ma poradove cislo 1. Variace cislo 4 ma poradove cislo 55.
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int greaterb4[15];
int num[15];
long long A( int a, int b )
{
long long res = 1;
for(int i = a; i > a-b; i--)
{
res *= i;
}
return res;
}
int main()
{
int kase;
cin >> kase;
int n, k;
for(int step = 1; step <= kase; step++)
{
memset( greaterb4, 0, sizeof greaterb4 );
cin >> n >> k;
for(int i = 1; i <= k; i++)
{
cin >> num[i];
for(int j = 1; j < i; j++)
{
if(num[j]>num[i]) greaterb4[i]++; //统计前面有几个更大的数
}
}//输入
long long ans = 0;
for(int i = 1; i <= k; i++)
{
long long tem = num[i]-(i-greaterb4[i]); //C(num,1) 其中num表示比这个数小且前面还没用过的数有多少个
tem *= A( n - i, k - i ); //剩下的数全排列
ans += tem;
}
printf( "Variace cislo %d ma poradove cislo %lld.\n", step, ans + 1 );
}
return 0;
}#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
const int MAXN = 1e6;
double f[MAXN];
void fac()
{
f[0] = 0;
for(int i = 1; i < MAXN; i++)
{
f[i] = f[i - 1] + log( i*1.0 ) ;
}
}
double getA( int n, int m ) //计算A(n,m) 的值，n为右下角的数，m为右上角的数
{
return f[n] - f[n - m];
}
double getC( int n, int m )//计算C(n,m) 的值，n为右下角的数，m为右上角的数
{
return f[n] - f[m] - f[n - m];
}
int main()
{
fac();
int n, m;
while(cin >> n >> m)
{
printf( "A(%d,%d) = %lf\n", n, m, exp( getA( n, m ) ) );
printf( "C(%d,%d) = %lf\n", n, m, exp( getC( n, m ) ) );
}
return 0;
}标签：soj1942 foto 数论 排列组合 快速排列组合
原文地址：http://blog.csdn.net/maxichu/article/details/45566039
