codeforces626F

CF626F Group Projects

 There are n students, each student has a capacity value ai. Now, we students into some groups (an arbitrary number), "disharmony degree" of each set value is the difference between the minimum capacity of the set of student ability value and the maximum capability value students. All of disharmony seek does not exceed the total number of program packets k.

Sample input and output

Input # 1copy

3 2
2 4 5

Output # 1copy

3

Input # 2copy

4 3
7 8 9 10

Output # 2copy

13

Input # 3copy

4 0
5 10 20 21

Output # 3copy

1

hint:n<=200,k<=1000

sol：

The sort a
note must be a degree of dissonance end value - the value of the starting point, there may be several intermediate non-final non-starting point, you can easily put
dp [i, j, k] i represents the number of front, there are a starting point not match j, k is the number of current total contribution to a program
transfer is made to enumerate the current point starting point, intermediate point and end point

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then it will hang badly, there is something a very pit, it is both a starting point and a point is the end

 

/ * 
The sort a 
note must be a degree of dissonance end value - the value of the starting point, there may be several intermediate non-final non-starting point, can easily put 
dp [i, j, k] denotes the i-th former number, as well as a starting point not match j, k is the number of current total contribution to a program 
transfer is made to enumerate the current point starting point, intermediate point, end point 
and then will hang badly, there is something a very pit, is both a point the starting point is the end point 
* / 
#include <bits / STDC ++ H.>
 the using  namespace STD; 
typedef int LL; 
inline LL Read () 
{ 
    LL S = 0 ; BOOL F = 0 ; char CH = '  ' ;
     the while (isdigit (! CH)) {F | = (CH == ' - ' ); CH = getchar ();}
     the while (isdigit (CH)) = {S (S <<3)+(s<<1)+(ch^48); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0) {putchar('-'); x=-x;}
    if(x<10) {putchar(x+'0'); return;}
    write(x/10); putchar((x%10)+'0');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int Mod=1000000007,N=205;
int n,m,a[N];
int dp[2][N][1005];
inline void Ad(int &x,int y)
{
    x+=y; x-=(x>=Mod)?Mod:0;
}
int main()
{
    freopen("codeforces626F_data.in","r",stdin);
    int i,j,k,t;
    R(n); R(m);
    for(i=1;i<=n;i++) R(a[i]);
    sort(a+1,a+n+1);
    dp[t=0][0][0]=1;
    for(i=1;i<=n;i++)
    {
        t^=1;
        for(j=0;j<=min(i,n>>1);j++)
        {
            int oo=a[i]-a[i-1];
            for(k=0;k<=m;k++)
            {
                dp[t][j][k]=0;
                if(j&&k>=(j-1)*oo) Ad(dp[t][j][k],dp[t^1][j-1][k-(j-1)*oo]);
                if(k>=j*oo) Ad(dp[t][j][k],1LL*dp[t^1][j][k-j*oo]*j%Mod);
                if(k>=(j+1)*oo) Ad(dp[t][j][k],1LL*dp[t^1][j+1][k-(j+1)*oo]*(j+1)%Mod);
                if(k>=j*oo) Ad(dp[t][j][k],dp[t^1][j][k-j*oo]);
            }
        }
    }
    int ans=0;
    for(i=0;i<=m;i++) Ad(ans,dp[t][0][i]);
    Wl(ans);
    return 0;
}

View Code