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\n",an2-len-an);}return 0;}