Balanced Number 数位dp

题意：　　给出求ab之间有多少个平衡数   4139为平衡数   以3为轴   1*1+4*2==9*1 

思路很好想但是一直wa  ：

注意要减去前导零的情况 0 00 000 0000   不能反复计算

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
#define N 3700+5
#define MID N/2
ll dp[20][20][N];
ll a[20];
ll dfs(int pos,int t,int state,bool lead,bool limit  )
{
    if(!pos)return state==0;

    if(!limit&&!lead&&dp[pos][t][state+MID]!=-1)return dp[pos][t][state+MID];
    ll ans=0;
    int up=limit?a[pos]:9;
    rep(i,0,up)
    {
        ans+=dfs(pos-1,t,state+i*(pos-t), lead&&i==0,limit&&i==a[pos]);
    }
    if(!limit&&!lead)dp[pos][t][state+MID]=ans;
    return ans;
}

ll solve(ll x)
{
    int pos=0;
    ll ans=0;
    while(x)
    {
        a[++pos]=x%10;
        x/=10;
    }
    rep(i,1,pos)
    ans+=dfs(pos,i,0,true,true);
    return ans-pos+1;//去除前导零
}
int main()
{
    int cas;
    CLR(dp,-1);
    RI(cas);
    while(cas--)
    {
        ll a,b;
        cin>>a>>b;
        printf("%lld\n",solve(b)-solve(a-1));
    }
    return 0;
}

View Code

或者

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
#define N 3700+5
#define MID N/2
ll dp[20][20][N];
ll a[20];
ll dfs(int pos,int t,int state,bool lead,bool limit  )
{
    if(!pos)return !lead&&state==0;

    if(!limit&&!lead&&dp[pos][t][state+MID]!=-1)return dp[pos][t][state+MID];
    ll ans=0;
    int up=limit?a[pos]:9;
    rep(i,0,up)
    {
        ans+=dfs(pos-1,t,state+i*(pos-t), lead&&i==0,limit&&i==a[pos]);
    }
    if(!limit&&!lead)dp[pos][t][state+MID]=ans;
    return ans;
}

ll solve(ll x)
{
    if(x<0)return 0;
    if(x==0)return 1;
    int pos=0;
    ll ans=0;
    while(x)
    {
        a[++pos]=x%10;
        x/=10;
    }
    rep(i,1,pos)
    ans+=dfs(pos,i,0,true,true);
    return ans+1;//去除前导零
}
int main()
{
    int cas;
    CLR(dp,-1);
    RI(cas);
    while(cas--)
    {
        ll a,b;
        cin>>a>>b;
        printf("%lld\n",solve(b)-solve(a-1));
    }
    return 0;
}

View Code