Hdu3079Balanced Number數位dp

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枚舉支點，然後就搞，記錄之前的點的力矩和。

#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>#include<math.h>using namespace std;typedef long long LL;LL up[100];LL dp[30][30][2000];LL dfs(LL x,LL pos,LL pre,LL flag){    if(x<=0) return pre==0;    if(pre<0) return 0;    if(!flag&&~dp[x][pos][pre]) return dp[x][pos][pre];    LL limit= flag? up[x]: 9,ret= 0;    for(LL i = 0;i<= limit;i++){        ret+=dfs(x-1,pos,pre+(x - pos )* i ,flag&&i==limit) ;    }    return flag? ret : dp[x][pos][pre]= ret;}LL solve(LL x){    LL len=0;    while(x){        up[++len]=x%10;        x/=10;    }    LL ans=0;    for(LL i = 1;i<= len;i++)        ans+=dfs(len,i,0,1);    return ans-len ;}int main(){    LL Icase;LL n,m;    scanf("%d",&Icase);    memset(dp,-1,sizeof(dp));    while(Icase--){        cin>>n>>m;        cout<<solve(m) - solve(n-1) <<endl;    }    return 0;}