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Count the Polygons

Time Limit: 2000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

E

Count the Polygons Input: Standard Input Output: Standard Output

A polygon is a plane figure that is bounded by a closed path and composed of a finite sequence of straight line segments. These segments are called its edges, and the points where two edges meet are the polygon's vertices.

You have got a set of N sticks of various lengths. How many ways can you choose K sticks from this set and form a polygon with K sides by joining the end points.

Input

The first line of input is an integer T (T<100) that indicates the number of test cases. Each case starts with a line containing 2 positive integers N and K ( 3≤N≤30 & 3≤K≤N ). The next line contains N positive integers in the range [1, 2^31), which represents the lengths of the available sticks. The integers are separated by a single space.

Output

For each case, output the case number followed by the number of valid polygons that can be formed by picking K sticks from the given N sticks.

Sample Input                             Output for Sample Input

4 4 3 10 10 20 20 6 4 1 1 1 1 1 1 4 3 10 20 30 100000000 6 6 2 3 4 5 6 7

Case 1: 2 Case 2: 15 Case 3: 0 Case 4: 1

 1000ms：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
int a[40],k,maxa,n;
ll ans,C[40][40],sum[40];
void dfs(ll summ,int cnt,int x)
{
    if(cnt==k)
    {
        if(summ>a[maxa])ans++;
        return ;
    }
    if(summ+sum[k-cnt+x-1]-sum[x-1]>a[maxa])
    {
        ans+=C[maxa-x][k-cnt];
        return ;
    }
    if(summ+sum[maxa-1]-sum[maxa-k+cnt-1]<=a[maxa])return ;
    if(x==maxa)return;
    dfs(summ+a[x],cnt+1,x+1);
    if(maxa-x>k-cnt)
    dfs(summ,cnt,x+1);
}
void init()
{
    for(int i=1; i<=30; i++)
    {
        C[i][0]=C[i][i]=1;
        for(int j=1; j<i; j++)
            C[i][j]=C[i-1][j-1]+C[i-1][j];
    }
}
int main()
{
    int T,i,j;
    init();
    scanf("%d",&T);
    for(i=1; i<=T; i++)
    {
        ans=0;
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&k);
        for(j=0; j<n; j++)
            scanf("%d",&a[j]);
        sort(a,a+n);
        sum[0]=a[0];
        for(j=1; j<n; j++)sum[j]=sum[j-1]+a[j];
        k--;
        for(j=k; j<n; j++)
        {
            maxa=j;
            dfs(0,0,0);
        }
        printf("Case %d: %lld
",i,ans);
    }
}

 9ms：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
int a[40],k,maxa,n;
ll ans,C[40][40],sum[40];
void dfs(ll summ,int cnt,int x)
{
    if(cnt==k)
    {
        if(summ>a[maxa])ans++;
        return ;
    }
    if(x==0)return;
    if(summ+sum[k-cnt]>a[maxa])
    {
        ans+=C[x][k-cnt];
        return ;
    }
    if(summ+sum[x]-sum[x-k+cnt]<=a[maxa])return ;
    dfs(summ+a[x],cnt+1,x-1);
    if(x>k-cnt)
        dfs(summ,cnt,x-1);
}
void init()
{
    for(int i=1; i<=31; i++)
    {
        C[i][0]=C[i][i]=1;
        for(int j=1; j<i; j++)
            C[i][j]=C[i-1][j-1]+C[i-1][j];
    }
}
int main()
{
    int T,i,j;
    init();
    scanf("%d",&T);
    for(i=1; i<=T; i++)
    {
        ans=0;
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&k);
        for(j=1; j<=n; j++)
            scanf("%d",&a[j]);
        sort(a+1,a+n+1);
        sum[0]=0;
        for(j=1; j<=n; j++)sum[j]=sum[j-1]+a[j];
        k--;
        for(j=k+1; j<=n; j++)
        {
            maxa=j;
            dfs(0,0,j-1);
        }
        printf("Case %d: %lld
",i,ans);
    }
}