HDU 3709 && UVALive 5004 && ZOJ 3416 Balanced Number 数位dp

题意

定义平衡数为将支点放在某个数字上,杠杆两边(每个数乘它离支点的距离)能够平衡的数字,求[l,r]内平衡数的个数.

题解

明显数位dp.
令$dp[i][j][sum]$表示$i$位的数字中,平衡点在第$j$个数字,平衡点左边的力矩减右边力矩的大小为$sum$的数字的个数.
枚举平衡点的位置,然后爆搞一下就完事了.

#include //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) x=-x,pc('-');
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
ll dp[25][20][1990];
int p,bit[25];
/*当前到第几位,支点位置,当前力矩差,最高位是否有要求*/
ll dfs(int now,int pit,int sum,int lim){
if (!now) return !sum; // 数位为0,看看杠杆是否平衡(力矩差为0)
if (!lim&&~dp[now][pit][sum]) return dp[now][pit][sum]; // 记忆化
ll zxy=0; int i,j,da=lim?bit[now]:9; 
for (i=0;i<=da;++i) zxy+=dfs(now-1,pit,sum+i*(now-pit),lim&&i==da); // 加上当前那一位对力矩差的贡献.
return lim?zxy:dp[now][pit][sum]=zxy; // 根据lim判断是否记录答案.
}

ll get(ll x){
ll llx=0;
for (p=0;x;x/=10) bit[++p]=x%10;
for (int i=1;i<=p;++i) llx+=dfs(p,i,0,1);
return llx-p;
}

int main(){
memset(dp,-1,sizeof dp);
for (int t=read();t--;){
  ll l,r; read(l),read(r);
  write(get(r)-get(l-1)),pl;
  }
}

谢谢大家.