Count Scorecards
Problem Statement :
In a tournament, n players play against each other exactly once. Each game results in exactly one player winning. There are no ties. You have been given a scorecard containing the scores of each player at the end of the tournament. The score of a player is the total number of games the player won in the tournament. However, the scores of some players might have been erased from the scorecard. How many possible scorecards are consistent with the input scorecard? Input Format The first line contains a single integer t denoting the number of test cases. t test cases follow. The first line of each test case contains a single integer n. The second line contains n space-separated integers s1,s2,...,sn. si denotes the score of the ith player. If the score of the ith player has been erased, it is represented by -1. Constraints 1 <= t <= 20 1 <= n <= 40 -1 <= si < n Output Format For each test case, output a single line containing the answer for that test case modulo 10^9+7.
Solution :
Solution in C :
In C++ :
#include<iostream>
#include<vector>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std ;
#define MAXN 42
#define MOD 1000000007
int n ;
vector<int> G ;
int check(vector<int> gg)
{
sort(gg.begin(),gg.end()) ;
int sum = 0 ;
for(int i = 0;i < n;i++)
{
sum += gg[i] ;
if(sum < i * (i + 1) / 2) return 0 ;
}
return 1 ;
}
int solve1(int k,int sum)
{
if(sum > n * (n - 1) / 2) return 0 ;
if(k == n) return sum == n * (n - 1) / 2 && check(G) ? 1 : 0 ;
if(G[k] != -1) return solve1(k + 1,sum + G[k]) ;
int ret = 0 ;
for(int i = 0;i < n;i++)
{
G[k] = i ;
ret += solve1(k + 1,sum + i) ;
G[k] = -1 ;
}
return ret ;
}
int solve1(vector<int> g)
{
G = g ;
n = G.size() ;
int ret = solve1(0,0) ;
return ret ;
}
int fac[MAXN],inv[MAXN] ;
int pow(int a,int b)
{
if(b == 0) return 1 ;
int ret = pow(a,b / 2) ;
ret = 1LL * ret * ret % MOD ;
if(b & 1) ret = 1LL * ret * a % MOD ;
return ret ;
}
int occ[MAXN],big[MAXN] ;
int memo[MAXN][MAXN][MAXN * MAXN] ;
int solve2(int k,int last,int sum)
{
if(last == n) return 0 ;
if(sum > n * (n - 1) / 2) return 0 ;
if(k == n) return big[last + 1] == 0 && sum == n * (n - 1) / 2 ? 1 : 0 ;
if(memo[k][last + 1][sum] != -1) return memo[k][last + 1][sum] ;
int occr = occ[last + 1] ;
int ret = 0,added = 0 ;
for(int i = 0;k + i <= n;i++)
{
if(sum + added < (k + i) * (k + i - 1) / 2) break ;
if(i >= occr) ret += 1LL * inv[i - occr] * solve2(k + i,last + 1,sum + added) % MOD ;
if(ret >= MOD) ret -= MOD ;
added += last + 1 ;
}
return memo[k][last + 1][sum] = ret ;
}
int solve2(vector<int> g)
{
fac[0] = inv[0] = 1 ;
for(int i = 1;i < MAXN;i++)
{
fac[i] = 1LL * i * fac[i - 1] % MOD ;
inv[i] = pow(fac[i],MOD - 2) ;
}
n = g.size() ;
int start = 0 ;
memset(occ,0,sizeof occ) ;
memset(big,0,sizeof big) ;
for(int i = 0;i < g.size();i++) if(g[i] != -1) occ[g[i]]++,start++ ;
for(int i = 0;i < n;i++)
for(int j = i;j < n;j++)
big[i] += occ[j] ;
memset(memo,255,sizeof memo) ;
int ret = solve2(0,-1,0) ;
ret = 1LL * ret * fac[n - start] % MOD ;
return ret ;
}
vector<int> gen()
{
int n = rand() % 10 + 1 ;
vector<int> ret ;
for(int i = 0;i < n;i++)
{
if(rand() % 2 == 0) ret.push_back(-1) ;
else ret.push_back(rand() % n) ;
}
return ret ;
}
void test()
{
for(int t = 1;t < 100;t++)
{
vector<int> g = gen() ;
int ret1 = solve1(g) ;
int ret2 = solve2(g) ;
cout << ret1 << " " << ret2 << endl ;
if(ret1 != ret2)
{
cout << "failed on: " << t << endl ;
cout << g.size() << " : " ;
for(int i = 0;i < g.size();i++) cout << g[i] << " " ; cout << endl ;
while(1) ;
}
}
}
void generate()
{
char in[] = "in .txt" ;
for(int test = 0;test < 10;test++)
{
in[2] = test + '0' ;
FILE * fout = fopen(in,"w") ;
int runs = 20 ;
fprintf(fout,"%d\n",runs) ;
for(int t = 0;t < runs;t++)
{
if(rand() % 3 != 0) n = rand() % 40 + 1 ;
else n = 40 - rand() % 10 ;
int per = 70 ;
vector<int> g ;
for(int i = 0;i < n;i++)
{
if(rand() % 100 < per) g.push_back(-1) ;
else g.push_back(rand() % n) ;
}
fprintf(fout,"%d\n",n) ;
for(int i = 0;i < n;i++)
{
if(i) fprintf(fout," ") ;
fprintf(fout,"%d",g[i]) ;
}
fprintf(fout,"\n") ;
}
}
}
int main()
{
// srand(time(NULL)) ;
// generate() ; return 0 ;
// test() ; return 0 ;
int runs ;
cin >> runs ;
while(runs--)
{
int n ;
vector<int> g ;
cin >> n ;
for(int i = 0;i < n;i++)
{
int k ;
cin >> k ;
g.push_back(k) ;
}
int ret = solve2(g) ;
printf("%d\n",ret) ;
}
return 0 ;
}
In C :
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
#include <stdio.h>
#include <stdlib.h>
#define P 1000000007
#define N 450
long long bin[50][50],jj,kk,h,ll,ii,t,a[50][50][1010],i,j,k,l,m,n,c[100],b[100];
int com(const void * xx, const void *yy)
{
if(*(long long *)xx < *(long long*)yy) return 1;
return -1;
}
int main()
{
for(i=0;i<50;i++) bin[i][0] = bin[i][i]=1;
for(i=1;i<50;i++)
for(j=1;j<i;j++) bin[i][j] = (bin[i-1][j-1]+ bin[i-1][j])%P;
scanf("%lld",&t);
while(t--)
{
scanf("%lld",&n);
for(i=0;i<n;i++) scanf("%lld",&c[i]);
qsort(c,n,sizeof(c[0]),com);
for(i=0;i<=n;i++) b[i]=0;
m=0;
for(i=0;i<n;i++) if(c[i]!=-1) b[c[i]]++; else m++;
for(i=0;i<=n;i++)
for(j=0;j<=n;j++)
for(k=0;k<=N;k++) a[i][j][k]=0;
a[n][0][0] = 1;
l=0;
for(i=n-1;i>=0;i--)
{
for(j=0;j<=m;j++)
{
h=0;
ll=l;
for(ii=0;ii<b[i];ii++)
{
h += (n-ll-j-1) - i;
ll++;
}
// printf("%lld %lld h %lld %lld=l\n",i,j,h,l);
for(k=0;k<=N;k++)
if(k+h>=0) a[i][j][k+h] = (a[i][j][k+h] + a[i+1][j][k])%P;
}
//for(ii=0;ii<=n;ii++)
// for(jj=0;jj<=m;jj++)
// for(kk=0;kk<=1000;kk++)
// if(a[ii][jj][kk]) printf("%lld %lld %lld -> %lld\n", ii,jj,kk,a[ii][jj][kk]);
//printf("stred -----------%lld\n",i);
// continue;
for(j=m;j>=0;j--)
for(k=0;k<N;k++)
{
h = k;
for(jj=1;jj<=j && h>=0;jj++)
{
h -= (n-l-b[i]-(j-jj)-1) - i;
if(h<0) break;
a[i][j][k] = (a[i][j][k] + bin[m-j+jj][jj]*a[i][j-jj][h])%P;
// if(a[i][j-jj][h]>0) printf("...%lld=i %lld=j %lld=k %lld=jj %lld=h\n",i,j,k,jj,h);
}
}
//for(ii=0;ii<=n;ii++)
// for(jj=0;jj<=m;jj++)
// for(kk=0;kk<=1000;kk++)
// if(a[ii][jj][kk]) printf("%lld %lld %lld -> %lld\n", ii,jj,kk,a[ii][jj][kk]);
//printf("-----------%lld\n",i);
l+=b[i];
}
/*
for(i=0;i<=n;i++)
for(j=0;j<=m;j++)
for(k=0;k<=1000;k++)
if(a[i][j][k]) printf("%lld %lld %lld -> %lld\n", i,j,k,a[i][j][k]);
for(i=0;i<=n;i++) printf("%lld %lld\n",i,c[i]);
*/
printf("%lld\n",(a[0][m][0]+P)%P);
//return 0;
}
return 0;
}
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