hdu_3709_Balanced Number(数位DP)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=3709

题意：给你一个区间，让你找平衡数的个数

题解：设dp[i][j][k]为前i位以第j位为支撑点的力矩和为k的方案数，注意的是0，00，000这些也是平衡数，所以要减掉一个len长度

#include<cstdio>
#include<cstring>
#define F(i,a,b) for(LL i=a;i<=b;i++)
typedef long long LL;

LL t,n,m,dp[20][20][2010],dig[20],len,an,an2,ii;

LL dfs(int pos,int now,int f,bool inf){
    if(!pos)return !f;
    if(f<0)return 0;
    if(!inf&&dp[pos][now][f]!=-1)return dp[pos][now][f];
    LL end=inf?dig[pos]:9,ans=0;
    F(i,0,end)ans+=dfs(pos-1,now,f+(pos-now)*i,inf&&(i==end));
    if(!inf)dp[pos][now][f]=ans;
    return ans;
}

int main(){
    memset(dp,-1,sizeof(dp));
    scanf("%lld",&t);
    while(t--){
        scanf("%lld%lld",&n,&m),n--;
        for(len=0;n;n/=10)dig[++len]=n%10;
        for(an=0,ii=len;ii>0;ii--)an+=dfs(len,ii,0,1);an-=len;
        for(len=0;m;m/=10)dig[++len]=m%10;
        for(an2=0,ii=len;ii>0;ii--)an2+=dfs(len,ii,0,1);
        printf("%lld
",an2-len-an);
    }
    return 0;
}