題意:給出一個數字,要求每位乘起來等於原數字的數字個數,每位的數字只能是2-9。
做法:可以發現2-9數字中素數只有2 3 5 7,而5和7是不能進行拆分和合並的,所以我們只需要考慮2和3,用dp[i][j][k]代表2的個數爲i個,3的個數爲j個,組成數字有k位的數字個數。遞推式也很簡單,枚舉最後一位爲2,3,4,6,8,9轉移即可。
再算出這個後,枚舉2和3組成的數字長度k,我們考慮放x個5進去,那麼就是(k+1)*(k+2)*.....*(k+x)/x!,放7也是一樣的。
比賽最後一小時纔看這個題。。結果因爲太蠢寫錯一個減枝,比賽結束後5分鐘突然醒悟刪了交就A了。。
AC代碼:
//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll long long
#define ull unsigned long long
#define eps 1e-8
#define NMAX 900000
#define MOD 1000000007
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
#define mp make_pair
template<class T>
inline void scan_d(T &ret)
{
char c;
int flag = 0;
ret=0;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c == '-')
{
flag = 1;
c = getchar();
}
while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
if(flag) ret = -ret;
}
ll dp[153][102][153];
char s[55];
int a[4];
ll dfs(int x, int y,int k)
{
if(x < 0 || y < 0 || k < 0) return 0;
if(x == 0 && y == 0 && k != 0) return dp[x][y][k] = 0;
if(dp[x][y][k] != -1) return dp[x][y][k];
if(k == 0 && x == 0 && y == 0) return dp[x][y][k] = 1;
ll &ans = dp[x][y][k];
ans = 0;
ans += (dfs(x-1,y,k-1)+dfs(x-2,y,k-1))%MOD;
ans %= MOD;
ans += (dfs(x-3,y,k-1)+dfs(x-1,y-1,k-1))%MOD;
ans %= MOD;
ans += (dfs(x,y-1,k-1)+dfs(x,y-2,k-1))%MOD;
ans %= MOD;
return ans;
}
ll cal(ll a, ll b)
{
ll ret = 1,tmp = a;
while(b)
{
if(b&1) ret = ret*tmp%MOD;
tmp = tmp*tmp%MOD;
b >>= 1;
}
return ret;
}
ll inv[155];
void init()
{
ll kao = 1;
inv[0] = 1;
for(int i = 1; i <= 152; i++)
{
kao *= (ll)i;
kao %= MOD;
inv[i] = cal(kao,MOD-2);
}
}
int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
// freopen("o1.txt","w",stdout);
#endif // GLQ
int len;
memset(dp,-1,sizeof(dp));
for(int i = 152; i >= 0; i--)
{
for(int j = 101; j >= 0; j--)
{
for(int k = 152; k >= 0; k--)
{
if(i == 0 && j == 0)
{
dp[i][j][k] = (k == 0)?1:0;
continue;
}
dfs(i,j,k);
}
}
}
init();
while(~scanf("%d",&len))
{
scanf("%s",s);
memset(a,0,sizeof(a));
for(int i = 0; i < len; i++)
{
int x = s[i]-'0';
if(x == 2) a[0]++;
if(x == 3) a[1]++;
if(x == 4) a[0] += 2;
if(x == 5) a[2]++;
if(x == 6)
{
a[0]++;
a[1]++;
}
if(x == 7) a[3]++;
if(x == 8) a[0] += 3;
if(x == 9) a[1] += 2;
}
ll ans1 = 0;
// cout<<dp[4][1][5]<<" "<<dp[1][0][1]<<endl;
for(int i = 1; i <= 152; i++)
{
ll tmp = dp[a[0]][a[1]][i],tmp2;
ll ans = 1;
if(tmp == 0) continue;
tmp2 = 1;
for(int j = i+1; j <= i+a[2]; j++)
{
tmp2 = tmp2*(ll)j;
tmp2 %= MOD;
}
ans = (tmp*tmp2%MOD*inv[a[2]]%MOD)%MOD;
tmp2 = 1;
for(int j = i+a[2]+1; j <= i+a[2]+a[3]; j++)
{
tmp2 = tmp2*(ll)j;
tmp2 %= MOD;
}
ans = ans*tmp2%MOD*inv[a[3]]%MOD;
// cout<<ans<<" "<<inv[a[3]]<<" "<<tmp2<<" "<<ans<<endl;
ans1 += ans;
ans1 %= MOD;
}
printf("%lld\n",ans1);
}
return 0;
}